media in matrix

Author: sachuverma

DSA Crack Sheet/README.md

@@ -45,6 +45,7 @@
 
 - [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)  
 - []( "view question") - [Cpp Solution](./solutions/.cpp)  
 
 ### Strings

DSA Crack Sheet/solutions/Median in a row-wise sorted Matrix.cpp

@@ -0,0 +1,62 @@
+/*
+Median in a row-wise sorted Matrix
+====================================
+
+Given a row wise sorted matrix of size RxC where R and C are always odd, find the median of the matrix.
+
+Example 1:
+Input:
+R = 3, C = 3
+M = [[1, 3, 5], 
+     [2, 6, 9], 
+     [3, 6, 9]]
+
+Output: 5
+Explanation:
+Sorting matrix elements gives us 
+{1,2,3,3,5,6,6,9,9}. Hence, 5 is median. 
+
+Example 2:
+Input:
+R = 3, C = 1
+M = [[1], [2], [3]]
+Output: 2
+Your Task:  
+You don't need to read input or print anything. Your task is to complete the function median() which takes the integers R and C along with the 2D matrix as input parameters and returns the median of the matrix.
+
+Expected Time Complexity: O(32 * R * log(C))
+Expected Auxiliary Space: O(1)
+
+Constraints:
+1<= R,C <=150
+1<= matrix[i][j] <=1000
+*/
+
+int median(vector<vector<int>> &m, int r, int c)
+{
+  int min = INT_MAX, max = INT_MIN;
+  for (int i = 0; i < r; i++)
+  {
+    if (m[i][0] < min)
+      min = m[i][0];
+    if (m[i][c - 1] > max)
+      max = m[i][c - 1];
+  }
+
+  int desired = (r * c + 1) / 2;
+  while (min < max)
+  {
+    int mid = min + (max - min) / 2;
+    int place = 0;
+
+    for (int i = 0; i < r; ++i)
+    {
+      place += upper_bound(m[i].begin(), m[i].end(), mid) - m[i].begin();
+    }
+    if (place < desired)
+      min = mid + 1;
+    else
+      max = mid;
+  }
+  return min;
+}