Add Max Selling Gap problem

Author: sahilbansal17

Maximum Selling Gap/README.md

@@ -0,0 +1,21 @@
+# Max Selling Gap
+[Try the problem](https://www.interviewbit.com/problems/max-distance/)
+
+Given an array `A` of integers, find the maximum of `j - i` subjected to the constraint of `A[i] <= A[j]`.
+
+If there is no solution possible, return `-1`.
+
+### Example 1
+
+```
+Input: [3,5,4,2]
+Output: 2
+Explanation: For the pair (3, 4).
+```
+
+### Example 2
+```
+Input: [5,7,1,9,6,3,2]
+Output: 4
+Explanation: For the pair (1, 2). There could be other answers but they'll have the same maximum gap.
+```

Maximum Selling Gap/solution.cpp

@@ -1,5 +1,5 @@
 #include <vector>
-
+#include <iostream>
 using namespace std;
 
 int maxSellingGap(const vector<int>& price) {
@@ -13,34 +13,31 @@ int maxSellingGap(const vector<int>& price) {
     }
 
     int left = 0;
-    int right = num - 1;
+    int right = 0;
 
     int maxGap = 0;
-    int minPrefix = -1;
-    while (left < num) {
+    int minPrefix = price[0];
+    while (left < num && right < num) {
         int elem = price[left];
-        // 5 7 1 9 6 3 2
-        // 9 9 9 9 6 3 2
-        //         ^
-
-        if (minPrefix == -1) {
-            // first iteration
-            while (suffix_max[right] < elem) {
-                --right;
-            }
-        } else if (minPrefix > elem) {
-            // shift the right to right
-            while (right < num && suffix_max[right] < elem) {
-                ++right;
-            }
-        }
-        minPrefix = min(minPrefix, elem);
+        //      5 7 1 9 6 3 2
+        //      5 5 1 1 1 1 1 (minPrefix)
+        //      9 9 9 9 6 3 2 (suffix_max)
+        //          ^
 
-        if (right < num && price[right] >= elem) {
+        minPrefix = min(minPrefix, elem);
+        if (minPrefix <= suffix_max[right]) {
             maxGap = max(maxGap, right - left);
+            ++right;
+        } else {
+            ++left;
         }
-        ++left;
     }
 
     return maxGap;
+}
+
+int main () {
+    vector<int> price = {5,7,1,9,6,3,2};
+    cout << maxSellingGap(price) << endl;
+    return 0;
 }

Maximum Selling Gap/solution.md

@@ -0,0 +1,26 @@
+# Solution
+
+<details>
+<summary> 
+Hint 1
+</summary>
+The brute force O(N^2) solution is obvious. Try to think if you can use sorting (you might have to also store the index information) to optimize it to O(N log N).
+</details>
+
+<details>
+<summary>
+Hint 2
+</summary>
+Think if you can further reduce the complexity of your algorithm to O(N). You might need extra space.
+Given a index i, we basically want to find the maximum index j, such that A[i] <= A[j].
+So, does storing the suffixMax (i.e. maximum starting from a given index till the end) help?
+
+At an index, if the suffixMax at next index is greater than it, definitely we can get a possible value from here. How to get the maximum one?
+</details>
+
+<details>
+<summary>
+Hint 3
+</summary>
+Now, also think if storing or calculating (while iterating) the prefixMin can help in some way!
+</details>

README.md

@@ -22,6 +22,7 @@ This repository contains the coding interview problems along with solutions.
 | [2 Keys Keyboard](2%20Keys%20Keyboard) | [text](2%20Keys%20Keyboard/solution.txt) | [CPP](2%20Keys%20Keyboard/solution.cpp) | DP, Maths | Medium |
 | [Very Hard Queries](Very%20Hard%20Queries) | [Solution article](Very%20Hard%20Queries/solution.md) | [CPP](Very%20Hard%20Queries/solution.cpp) | Queries, Maths | Medium |
 | [Deterministic Finite Automaton](Deterministic%20Finite%20Automaton/) | [Solution article](Deterministic%20Finite%20Automaton/solution.md) | [CPP](Deterministic%20Finite%20Automaton/solution.cpp) | Recursion, Dynamic Programming | Medium |
+| [Maximum Selling Gap](Maximum%20Selling%20Gap) | [Solution article](Maximum%20Selling%20Gap/solution.md) | [CPP](Maximum%20Selling%20Gap/solution.cpp) | Arrays | Medium |
 
 # CS Fundamentals