row with max 1

Author: sachuverma

Diff for: DSA Crack Sheet/README.md

@@ -46,6 +46,7 @@
 - [Spirally traversing a matrix](https://practice.geeksforgeeks.org/problems/spirally-traversing-a-matrix-1587115621/1# "view question") - [Cpp Solution](./solutions/Spirally%20traversing%20a%20matrix.cpp)  
 - [Search a 2D Matrix](https://leetcode.com/problems/search-a-2d-matrix/ "view question") - [Cpp Solution](./solutions/Search%20a%202D%20Matrix.cpp)  
 - [Median in a row-wise sorted Matrix](https://practice.geeksforgeeks.org/problems/median-in-a-row-wise-sorted-matrix1527/1 "view question") - [Cpp Solution](./solutions/Median%20in%20a%20row-wise%20sorted%20Matrix.cpp)  
+- [Row with max 1s](https://practice.geeksforgeeks.org/problems/row-with-max-1s0023/1# "view question") - [Cpp Solution](./solutions/Row%20with%20max%201s.cpp)  
 - []( "view question") - [Cpp Solution](./solutions/.cpp)  
 
 ### Strings

Diff for: DSA Crack Sheet/solutions/Row with max 1s.cpp

@@ -0,0 +1,53 @@
+/*
+Row with max 1s
+===============
+
+Given a boolean 2D array of n x m dimensions where each row is sorted. Find the 0-based index of the first row that has the maximum number of 1's.
+
+Example 1:
+Input: 
+N = 4 , M = 4
+Arr[][] = {{0, 1, 1, 1},
+           {0, 0, 1, 1},
+           {1, 1, 1, 1},
+           {0, 0, 0, 0}}
+Output: 2
+Explanation: Row 2 contains 4 1's (0-based
+indexing).
+
+Example 2:
+Input: 
+N = 2, M = 2
+Arr[][] = {{0, 0}, {1, 1}}
+Output: 1
+Explanation: Row 1 contains 2 1's (0-based
+indexing).
+
+Your Task:  
+You don't need to read input or print anything. Your task is to complete the function rowWithMax1s() which takes the array of booleans arr[][], n and m as input parameters and returns the 0-based index of the first row that has the most number of 1s. If no such row exists, return -1.
+
+Expected Time Complexity: O(N+M)
+Expected Auxiliary Space: O(1)
+
+Constraints:
+1 ≤ N, M ≤ 103
+0 ≤ Arr[i][j] ≤ 1 
+*/
+
+int rowWithMax1s(vector<vector<int>> arr, int R, int C)
+{
+  int max_row_index = -1;
+  int j = upper_bound(arr[0].begin(), arr[0].end(), 0) - arr[0].begin();
+  if (j == 0)
+    return 0;
+
+  for (int i = 1; i < R; i++)
+  {
+    while (j - 1 >= 0 && arr[i][j - 1] == 1)
+    {
+      j = j - 1;
+      max_row_index = i;
+    }
+  }
+  return max_row_index;
+}