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‎README.md

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# Java Placement Preparation DSA CRACKER SHEET 💻🦸‍♂️🐱‍👤[357/450]
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# Java Placement Preparation DSA CRACKER SHEET 💻🦸‍♂️🐱‍👤[390/450]
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☄ This is a full fled-ged repository for learning Java Language & DSA for Placement Preparation.
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‎Relevel_Test.java

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/*
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Problem -> Solve History Ranking
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DUMB JERRY
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DUMB JERRY
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Problem Statement
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As usual, Jerry and Tom are fighting. Tom wants to prove that Jerry is Dumb, So he made a problem and asked jerry.
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Given a sequence A of size N, and an operation.
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In one operation you can pick an element either from the front or back of the Sequence and write it down and delete it from the array.
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Jerry has to perform the given operation until sequence A becomes empty.
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Tom wants the written sequence to be lexicographically maximum possible by performing the given operation on sequence A.
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Jerry can’t solve the problem on his own. So you have to help him.
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Note:-
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Sequence X of length N is lexicographically greater than Sequence Y of length N, if there exist such i (1<=i<=N), that Xi >Yi and for any j (1<=j<i) Xi=Yi.
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Constraint
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• 1 <= N <= 105
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• 0 <= Ai <= 109
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• All input values are integer.
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Input Format
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• First-line contains N, length of the sequence A
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• Second line contains N space-separated integers denoting the sequence A.
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Output Format
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• In first print the lexicographically maximum sequence made by performing operations on sequence A
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Sample Input 1
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3
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3 1 2
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Sample Output 1
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3 2 1
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Explanation of Sample 1
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For test case1
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We get sequence {3,1,2} if we
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Pick the element from the front of the sequence (3) and delete it from the original sequence {1,2}
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Pick the element from the front of the sequence (3,1) and delete it from the original sequence {2}
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Pick the element from the front of the sequence (3,1,2) and delete it from the original sequence {}
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We get sequence {3,2,1} if we
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Pick the element from the front of the sequence (3) and delete it from the original sequence {1,2}
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Pick the element from the back of the sequence (3,2) and delete it from the original sequence {1}
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Pick the element from the front of the sequence (3,2,1) and delete it from the original sequence {}
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We get sequence {2,3,1} if we
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Pick the element from the back of the sequence (2) and delete it from the original sequence {3,1}
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Pick the element from the front of the sequence (2,3) and delete it from the original sequence {1}
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Pick the element from the front of the sequence (2,3,1) and delete it from the original sequence {}
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We get sequence {2,1,3} if we
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Pick the element from the back of the sequence (2) and delete it from the original sequence {3,1}
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Pick the element from the back of the sequence (2,1) and delete it from the original sequence {3}
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Pick the element from the front of the sequence (2,1,3) and delete it from the original sequence {}
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Compare to other {3,2,1} is lexicographically largest.
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Things to Note for the Candidate
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You can use your own IDE like Visual Studio Code, Eclipse, or any other IDE that you are comfortable with to build your solution code.
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The IDE provided on the platform is purely for submission. Avoid using the IDE for coding out the solution.
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Test against any custom input in your own IDE. In the IDE provided on the platform, you cannot pass custom test cases and see the output.
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Use standard input and standard output in your program for the IDE to run the test cases smoothly against your code. We are giving a sample problem statement along with a solution that will pass the test cases in the IDE. - Sample Problem Statement (Right Click and Open in New Tab to view.)
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*/
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/* package whatever; // don't place package name! */
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import java.util.*;
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import java.lang.*;
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import java.io.*;
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/* Name of the class has to be "Main" only if the class is public. */
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class Relevel_Test
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{
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public static void main (String[] args) throws java.lang.Exception
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{
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// your code goes here
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Scanner sc = new Scanner(System.in);
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int m = sc.nextInt();
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int n = sc.nextInt();
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int[] x = new int[m];
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int[] y = new int[n];
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for (int a : x) {
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x[a] = sc.nextInt();
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}
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for (int b : y) {
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y[b] = sc.nextInt();
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}
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}
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public static int solve() {
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}
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}
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//
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///*
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//Problem -> Solve History Ranking
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//
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//DUMB JERRY
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//DUMB JERRY
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//
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//Problem Statement
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//
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//As usual, Jerry and Tom are fighting. Tom wants to prove that Jerry is Dumb, So he made a problem and asked jerry.
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//
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//Given a sequence A of size N, and an operation.
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//
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//In one operation you can pick an element either from the front or back of the Sequence and write it down and delete it from the array.
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//Jerry has to perform the given operation until sequence A becomes empty.
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//
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//Tom wants the written sequence to be lexicographically maximum possible by performing the given operation on sequence A.
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//
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//Jerry can’t solve the problem on his own. So you have to help him.
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//
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//Note:-
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//
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//Sequence X of length N is lexicographically greater than Sequence Y of length N, if there exist such i (1<=i<=N), that Xi >Yi and for any j (1<=j<i) Xi=Yi.
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//
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//Constraint
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//
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//• 1 <= N <= 105
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//• 0 <= Ai <= 109
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//
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//• All input values are integer.
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//
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//Input Format
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//
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//• First-line contains N, length of the sequence A
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//
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//• Second line contains N space-separated integers denoting the sequence A.
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//
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//Output Format
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//
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//• In first print the lexicographically maximum sequence made by performing operations on sequence A
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//
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//Sample Input 1
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//
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//3
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//
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//3 1 2
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//
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//Sample Output 1
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//
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//3 2 1
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//
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//Explanation of Sample 1
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//
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//For test case1
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//
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//We get sequence {3,1,2} if we
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//Pick the element from the front of the sequence (3) and delete it from the original sequence {1,2}
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//Pick the element from the front of the sequence (3,1) and delete it from the original sequence {2}
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//Pick the element from the front of the sequence (3,1,2) and delete it from the original sequence {}
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//We get sequence {3,2,1} if we
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//Pick the element from the front of the sequence (3) and delete it from the original sequence {1,2}
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//Pick the element from the back of the sequence (3,2) and delete it from the original sequence {1}
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//Pick the element from the front of the sequence (3,2,1) and delete it from the original sequence {}
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//We get sequence {2,3,1} if we
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//Pick the element from the back of the sequence (2) and delete it from the original sequence {3,1}
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//Pick the element from the front of the sequence (2,3) and delete it from the original sequence {1}
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//Pick the element from the front of the sequence (2,3,1) and delete it from the original sequence {}
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//We get sequence {2,1,3} if we
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//Pick the element from the back of the sequence (2) and delete it from the original sequence {3,1}
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//Pick the element from the back of the sequence (2,1) and delete it from the original sequence {3}
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//Pick the element from the front of the sequence (2,1,3) and delete it from the original sequence {}
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//Compare to other {3,2,1} is lexicographically largest.
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//
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//Things to Note for the Candidate
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//
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//You can use your own IDE like Visual Studio Code, Eclipse, or any other IDE that you are comfortable with to build your solution code.
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//The IDE provided on the platform is purely for submission. Avoid using the IDE for coding out the solution.
78+
//Test against any custom input in your own IDE. In the IDE provided on the platform, you cannot pass custom test cases and see the output.
79+
//Use standard input and standard output in your program for the IDE to run the test cases smoothly against your code. We are giving a sample problem statement along with a solution that will pass the test cases in the IDE. - Sample Problem Statement (Right Click and Open in New Tab to view.)
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//
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//
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//
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// */
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//
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//
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///* package whatever; // don't place package name! */
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//
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//import java.util.*;
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//import java.lang.*;
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//import java.io.*;
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//
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///* Name of the class has to be "Main" only if the class is public. */
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//class Relevel_Test
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//{
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//
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// public static void main (String[] args) throws java.lang.Exception
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// {
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// // your code goes here
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// Scanner sc = new Scanner(System.in);
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// int m = sc.nextInt();
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// int n = sc.nextInt();
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//
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// int[] x = new int[m];
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// int[] y = new int[n];
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//
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// for (int a : x) {
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// x[a] = sc.nextInt();
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// }
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//
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// for (int b : y) {
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// y[b] = sc.nextInt();
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// }
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// }
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//
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// public static int solve() {
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//
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// }
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//}

‎arrays/2DArray.java

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// 2d array 0 1 , sorted in asc order
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// min number of zeroes
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/**
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* *
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* [0 0 0 0 0], -> 5
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* [0 0 0 0 0], -> 5
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* [0 1 1 1 1], -> 1
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* [1 1 1 1 1], -> 0
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*
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* [0 0 0 0 0 0 0 0 0 0 1 1 1 1 1 1 1 1 1]
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*
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* O(m.log n)
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*/
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package arrays;
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public class
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//// 2d array 0 1 , sorted in asc order
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//// min number of zeroes
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//
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///**
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// * *
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// * [0 0 0 0 0], -> 5
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// * [0 0 0 0 0], -> 5
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// * [0 1 1 1 1], -> 1
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// * [1 1 1 1 1], -> 0
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// *
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// * [0 0 0 0 0 0 0 0 0 0 1 1 1 1 1 1 1 1 1]
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// *
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// * O(m.log n)
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// */
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//
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//package arrays;
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//
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//public class
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//

‎arrays/interviewProblems/Aggressive_Cows.java

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package arrays.interviewProblems;
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import java.util.*;
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// Website -> CodeStudio
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// Problem Name -> Aggressive Cows
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public class Aggressive_Cows
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{
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public static boolean isPossible(ArrayList<Integer> stalls, int k, int mid) {
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int cowCount = 1;
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int lastPos = stalls.get(0);
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for(int i = 0; i < stalls.size(); i++) {
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if(stalls.get(i) - lastPos >= mid) {
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cowCount++;
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if(cowCount == k) {
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return true;
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}
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lastPos = stalls.get(i);
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}
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}
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return false;
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}
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public static int aggressiveCows(ArrayList<Integer> stalls, int k)
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{
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// Write your code here.
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Collections.sort(stalls);
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int start = 0, max = -1;
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for(int i = 0; i < stalls.size(); i++) {
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max = Math.max(max, stalls.get(i));
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}
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int end = max;
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int ans = -1;
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int mid = start + (end - start) / 2;
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while(start <= end) {
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if(isPossible(stalls, k, mid)) {
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ans = mid;
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start = mid + 1;
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}
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else
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end = mid - 1;
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mid = start + (end - start) / 2;
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}
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return ans;
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}
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public static void main(String[] args) {
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}
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}

‎out/production/Java-Programs/.cph/.Append.java_7d2b8bb570ab4d08b71fd1cd68eb8af0.prob renamed to ‎arrays/interviewProblems/CandyDistribution.java

File renamed without changes.

‎arrays/interviewProblems/FindMinOps.java

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package arrays;
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/* Problem Title :-> Minimum no. of operations required to make an arrays.array palindrome */
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public class Array_Problem_35 {
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static int findMinOps(int[] a, int n) {
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//Initialize result
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int ans = 0;
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//Start from two corners
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for(int i=0,j=n-1;i<=j;) {
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if(a[i] == a[j]) {
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i++;
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j--;
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}else if(a[i] > a[j]) {
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j--;
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a[j] += a[j+1];
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ans++;
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}else {
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i++;
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a[i] += a[i-1];
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ans++;
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}
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}
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return ans;
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}
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public static void main(String[] args) {
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int[] a = new int[] {1,2,3,4,5,6,7,8,9,10};
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System.out.println("Count of minimum operations is " + findMinOps(a , a.length));
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}
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}

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