Search a 2D Matrix in C++

Suppose we have write an efficient algorithm to searches for a value in one m x n matrix. This matrix has some properties like below −

• Each row is sorted from left to right
• The first number of each row is greater than the last integer of the previous row.

So if the matrix is like −

And if the target value is 16, then the output will be True.

Let us see the steps −

• n := number of rows, if n is 0, then return false, m := number of columns, if m = 0, then return false
• low := 0 and high := n – 1
• while low < high
  • mid := low + (high – low + 1)/2
  • if mat[mid, 0] <= target, then low := mid, otherwise high := mid – 1
• rlow := 0 and rhigh := m – 1 and ans := 0
• while rlow <= rhigh
  • mid := rlow + (rhigh - rlow)/2
  • if mat[low, mid] = target, then ans := 1, and break the loop
  • otherwise when matrix[low, mid] < target, then rlow := mid + 1
  • else rhigh := mid – 1
• return ans

Let us see the following implementation to get better understanding −

Example

 Live Demo

#include <bits/stdc++.h>
using namespace std;
typedef long long int lli;
class Solution {
   public:
   bool searchMatrix(vector<vector<int>>& matrix, int target) {
      lli n,m;
      n = matrix.size();
      if(!n)return false;
      m = matrix[0].size();
      if(!m)return false;
      lli low = 0, high = n-1;
      while(low<high){
         lli mid = low + ( high - low +1)/2;
         if(matrix[mid][0]<=target)low = mid;
         else high = mid -1;
      }
      lli rlow = 0, rhigh = m-1;
      lli ans = 0;
      while(rlow<=rhigh){
         lli mid = rlow+(rhigh - rlow)/2;
         if(matrix[low][mid] == target){
            ans =1;
            break;
         }else if(matrix[low][mid]<target)rlow=mid+1;
         else rhigh= mid-1;
      }
      return ans;
   }
};
main(){
   Solution ob;
   vector<vector<int>> v = {{1,3,5,7},{10,11,16,20},{23,30,34,50},{53,62,78,98}};
   cout << ob.searchMatrix(v, 16);
}

Input

[[1,3,5,7],[10,11,16,20],[23,30,34,50],[53,62,78,98]]
16

Output

1