rename folders

Author: akgmage

2D_Arrays/2d_binary_search.cpp renamed to 2D Arrays/2d_binary_search.cpp

@@ -1,48 +1,48 @@
-"""
-    Intuition:
-        If you have to guess a magic number from 1-100, the best first attempt would be to guess '50'
-        or in other words, the middle. If I tell you that the magic number is higher,
-        you now don't need to consider all numbers 1-50,
-        and if it is lower you wouldn't need to consider numbers 50-100!!
-
-        In Binary Search, we follow the same idea,
-        1. Compar the target with the middle element.
-        2. If the target is higher, then the target can only lie in the right (greater) subarray. We re-calculate mid and repeat step 1.
-        3. If the target is lower, the target  can only lie in the left (lower) half. We re-calculate mid and repeat step 1.
-
-        Binary search can only operate on a sorted array.
-        Further reading: https://en.wikipedia.org/wiki/Binary_search_algorithm
-"""
-
-
-
-
-import math
-
-def binary_search(lst, target):
-    if not lst:
-        return -1
-    lo = 0
-    hi = len(lst)-1
-
-    while lo <= hi:
-        mid = math.floor(lo + (hi - lo) / 2)  # Find mid. math.floor is used to round floats down.
-        if lst[mid] < target:  # Element in mid is lower than target.
-            lo = mid + 1  # Our low (start) becomes the element after mid.
-        elif lst[mid] > target:  # Element in mid is higher than target.
-            hi = mid - 1  # Our high (end) becomes the element before mid.
-        elif lst[mid] == target:
-            print(f"Found {target} at index {mid}.")
-            return mid
-    print(f"Target {target} not found.")
-    return -1
-
-
-arr = [10, 20, 30, 50, 60, 80, 110, 130, 140, 170]
-binary_search(arr, 80)
-binary_search(arr, 10)
-binary_search(arr, 110)
-binary_search(arr, 20)
-binary_search(arr, 140)
-binary_search(arr, 2)
-binary_search(arr, 1)
+"""
+    Intuition:
+        If you have to guess a magic number from 1-100, the best first attempt would be to guess '50'
+        or in other words, the middle. If I tell you that the magic number is higher,
+        you now don't need to consider all numbers 1-50,
+        and if it is lower you wouldn't need to consider numbers 50-100!!
+
+        In Binary Search, we follow the same idea,
+        1. Compar the target with the middle element.
+        2. If the target is higher, then the target can only lie in the right (greater) subarray. We re-calculate mid and repeat step 1.
+        3. If the target is lower, the target  can only lie in the left (lower) half. We re-calculate mid and repeat step 1.
+
+        Binary search can only operate on a sorted array.
+        Further reading: https://en.wikipedia.org/wiki/Binary_search_algorithm
+"""
+
+
+
+
+import math
+
+def binary_search(lst, target):
+    if not lst:
+        return -1
+    lo = 0
+    hi = len(lst)-1
+
+    while lo <= hi:
+        mid = math.floor(lo + (hi - lo) / 2)  # Find mid. math.floor is used to round floats down.
+        if lst[mid] < target:  # Element in mid is lower than target.
+            lo = mid + 1  # Our low (start) becomes the element after mid.
+        elif lst[mid] > target:  # Element in mid is higher than target.
+            hi = mid - 1  # Our high (end) becomes the element before mid.
+        elif lst[mid] == target:
+            print(f"Found {target} at index {mid}.")
+            return mid
+    print(f"Target {target} not found.")
+    return -1
+
+
+arr = [10, 20, 30, 50, 60, 80, 110, 130, 140, 170]
+binary_search(arr, 80)
+binary_search(arr, 10)
+binary_search(arr, 110)
+binary_search(arr, 20)
+binary_search(arr, 140)
+binary_search(arr, 2)
+binary_search(arr, 1)

2D_Arrays/search_el.cpp renamed to 2D Arrays/search_el.cpp

@@ -1,18 +1,18 @@
-// Program to count the number of bits that are set to 1 in an integer
-// The following program tests bits one at a time starting with the least-significant bit.
-// Since we perform O(1) computation per bit, the time complexity is O(n) where n is number of bits in the integer
-// Best case time complexity is O(1), if the input io 0
-#include<bits/stdc++.h>
-using namespace std;
-int find_set_bits(int n){
-    int set_bits = 0;
-    while(n){
-        set_bits += (n & 1);
-        n >>= 1;
-    }
-    return set_bits;
-}
-int main(){
-    cout << find_set_bits(4) << endl;
-    return 0;
-}
+// Program to count the number of bits that are set to 1 in an integer
+// The following program tests bits one at a time starting with the least-significant bit.
+// Since we perform O(1) computation per bit, the time complexity is O(n) where n is number of bits in the integer
+// Best case time complexity is O(1), if the input io 0
+#include<bits/stdc++.h>
+using namespace std;
+int find_set_bits(int n){
+    int set_bits = 0;
+    while(n){
+        set_bits += (n & 1);
+        n >>= 1;
+    }
+    return set_bits;
+}
+int main(){
+    cout << find_set_bits(4) << endl;
+    return 0;
+}

Binary_Search/CeilLetter.java renamed to Binary Search/CeilLetter.java

@@ -1,52 +1,52 @@
-/*
-    Given two strings str1 and str2 and following three operations that can performed on str1. 
-    1) Insert
-    2) Remove
-    3) Replace
-    Find minimum number of operations required to convert ‘str1’ into ‘str2’. 
-    For example if input strings are CAT AND CAR the edit distance is 1.
-
-    Input  :  s1 : saturday  s2 : sunday 
-    Output :  3  
-*/
-// Dynamic Programming Solution : TC O(n^2)
-// Porgram Author : Abhisek Kumar Gupta
-#include<bits/stdc++.h>
-using namespace std;
-int find_edit_distance(string s1, string s2, int l1, int l2){ 
-    int dp[100][100] = {};
-    for(int i = 0; i <= l1; i++){
-        dp[i][0] = i;
-    }
-    for(int i = 0; i <= l2; i++){
-        dp[0][i] = i;
-    }
-    for(int i = 1; i <= l1; i++){
-        for(int j = 1; j <= l2; j++){
-            if(s1[i] == s2[j])
-                dp[i][j] = dp[i - 1][j - 1];
-            else{
-                int del = dp[i][j - 1];
-                int replace = dp[i - 1][j - 1];
-                int insert =  dp[i - 1][j];
-                dp[i][j] = min(del, min(replace, insert)) + 1;
-            }
-        }
-    }
-    for(int i = 0; i <= l1; i++){
-        for(int j = 0; j <= l2; j++){
-            cout << setw(5) <<dp[i][j] << " ";
-        }
-        cout << "\n";
-    }
-    return dp[l1][l2];
-}
-int main(){
-    string s1 = "abhisek";
-    string s2 = "tsunade";
-    int l1 = s1.length() - 1;
-    int l2 = s2.length() - 1;
-    int result = find_edit_distance(s1, s2, l1, l2);
-    cout << result;
-    return 0;
-}
+/*
+    Given two strings str1 and str2 and following three operations that can performed on str1. 
+    1) Insert
+    2) Remove
+    3) Replace
+    Find minimum number of operations required to convert ‘str1’ into ‘str2’. 
+    For example if input strings are CAT AND CAR the edit distance is 1.
+
+    Input  :  s1 : saturday  s2 : sunday 
+    Output :  3  
+*/
+// Dynamic Programming Solution : TC O(n^2)
+// Porgram Author : Abhisek Kumar Gupta
+#include<bits/stdc++.h>
+using namespace std;
+int find_edit_distance(string s1, string s2, int l1, int l2){ 
+    int dp[100][100] = {};
+    for(int i = 0; i <= l1; i++){
+        dp[i][0] = i;
+    }
+    for(int i = 0; i <= l2; i++){
+        dp[0][i] = i;
+    }
+    for(int i = 1; i <= l1; i++){
+        for(int j = 1; j <= l2; j++){
+            if(s1[i] == s2[j])
+                dp[i][j] = dp[i - 1][j - 1];
+            else{
+                int del = dp[i][j - 1];
+                int replace = dp[i - 1][j - 1];
+                int insert =  dp[i - 1][j];
+                dp[i][j] = min(del, min(replace, insert)) + 1;
+            }
+        }
+    }
+    for(int i = 0; i <= l1; i++){
+        for(int j = 0; j <= l2; j++){
+            cout << setw(5) <<dp[i][j] << " ";
+        }
+        cout << "\n";
+    }
+    return dp[l1][l2];
+}
+int main(){
+    string s1 = "abhisek";
+    string s2 = "tsunade";
+    int l1 = s1.length() - 1;
+    int l2 = s2.length() - 1;
+    int result = find_edit_distance(s1, s2, l1, l2);
+    cout << result;
+    return 0;
+}

Binary_Search/CeilOfTarget.java renamed to Binary Search/CeilOfTarget.java

@@ -1,44 +1,44 @@
-/*
-    Given two strings str1 and str2 and following three operations that can performed on str1. 
-    1) Insert
-    2) Remove
-    3) Replace
-    Find minimum number of operations required to convert ‘str1’ into ‘str2’. 
-    For example if input strings are CAT AND CAR the edit distance is 1.
-
-    Input  : s1 : saturday  s2 : sunday 
-    Output : 3  
-*/
-// Memoized Solution : TC O(n^2)
-// Porgram Author : Abhisek Kumar Gupta
-#include<bits/stdc++.h>
-using namespace std;
-int calls = 0;
-int memoize[1009][1009];
-int find_edit_distance(string s1, string s2, int l1, int l2){
-    calls++;
-    if(l1 == 0)
-        return l2;
-    if(l2 == 0)
-        return l1;
-    if(memoize[l1][l2] != -1) return memoize[l1][l2];          
-    if(s1[l1] == s2[l2]){
-        return find_edit_distance(s1, s2, l1 - 1, l2 - 1);
-    }
-    int del = find_edit_distance(s1, s2, l1, l2 - 1);
-    int replace = find_edit_distance(s1, s2, l1 - 1, l2 - 1);
-    int insert = find_edit_distance(s1, s2, l1 - 1, l2);
-    memoize[l1][l2] = min (del, min(replace, insert)) + 1;
-    return min (del, min(replace, insert)) + 1;
-}
-int main(){
-    memset(memoize, -1, sizeof(memoize));
-    string s1 = "abhisek";
-    string s2 = "tsunade";
-    int l1 = s1.length() - 1;
-    int l2 = s2.length() - 1;
-    int result = find_edit_distance(s1, s2, l1, l2);
-    cout << result;
-    cout << "\n" << calls;
-    return 0;
+/*
+    Given two strings str1 and str2 and following three operations that can performed on str1. 
+    1) Insert
+    2) Remove
+    3) Replace
+    Find minimum number of operations required to convert ‘str1’ into ‘str2’. 
+    For example if input strings are CAT AND CAR the edit distance is 1.
+
+    Input  : s1 : saturday  s2 : sunday 
+    Output : 3  
+*/
+// Memoized Solution : TC O(n^2)
+// Porgram Author : Abhisek Kumar Gupta
+#include<bits/stdc++.h>
+using namespace std;
+int calls = 0;
+int memoize[1009][1009];
+int find_edit_distance(string s1, string s2, int l1, int l2){
+    calls++;
+    if(l1 == 0)
+        return l2;
+    if(l2 == 0)
+        return l1;
+    if(memoize[l1][l2] != -1) return memoize[l1][l2];          
+    if(s1[l1] == s2[l2]){
+        return find_edit_distance(s1, s2, l1 - 1, l2 - 1);
+    }
+    int del = find_edit_distance(s1, s2, l1, l2 - 1);
+    int replace = find_edit_distance(s1, s2, l1 - 1, l2 - 1);
+    int insert = find_edit_distance(s1, s2, l1 - 1, l2);
+    memoize[l1][l2] = min (del, min(replace, insert)) + 1;
+    return min (del, min(replace, insert)) + 1;
+}
+int main(){
+    memset(memoize, -1, sizeof(memoize));
+    string s1 = "abhisek";
+    string s2 = "tsunade";
+    int l1 = s1.length() - 1;
+    int l2 = s2.length() - 1;
+    int result = find_edit_distance(s1, s2, l1, l2);
+    cout << result;
+    cout << "\n" << calls;
+    return 0;
 }