# 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 −

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

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