**Matrix Search Sequel - Amazon Top Interview Questions**

### Problem Statement :

Given a two-dimensional integer matrix, where every row and column is sorted in ascending order, return whether an integer target exists in the matrix. This should be done in \mathcal{O}(n + m)O(n+m) time. Constraints n, m ≤ 250 where n and m are the number of rows and columns in matrix Example 1 Input matrix = [ [1, 3, 9], [2, 5, 10], [5, 7, 13] ] target = 7 Output True

### Solution :

` ````
Solution in C++ :
bool solve(vector<vector<int>>& matrix, int target) {
int n = matrix.size();
int m = matrix[0].size();
int i = 0;
int j = m - 1;
while (i < n && j >= 0) {
if (matrix[i][j] == target)
return true;
else if (matrix[i][j] > target)
j--;
else
i++;
}
return false;
}
```

` ````
Solution in Java :
import java.util.*;
class Solution {
public boolean solve(int[][] matrix, int target) {
for (int row = 0; row < matrix.length; row++) {
int idx = Arrays.binarySearch(matrix[row], target);
if (idx < 0)
continue;
else
return true;
}
return false;
}
}
```

` ````
Solution in Python :
class Solution:
def found(self, row, target):
lo = 0
hi = len(row)
while lo <= hi:
mid = lo + (hi - lo) // 2
if row[mid] == target:
return True
elif target < row[mid]:
hi = mid - 1
else:
lo = mid + 1
return False
def solve(self, matrix, target):
for row in matrix:
if row[0] <= target and row[-1] >= target:
if self.found(row, target):
return True
return False
```

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