# One binary search to find the row of target value and another to find the target in that row!!!! O(log(row)+log(col))

• `````` bool searchMatrix(vector<vector<int>>& matrix, int target) {
int r=matrix.size() , c=matrix[0].size(),mid;
// condition to check target is in matrix range
if(target < matrix[0][0] || target > matrix[r-1][c-1]) return false;
//first binary search to find the row number in which the target is present
int s=0 , e=r-1;
while(s <= e){
mid =s+ (e-s)/2;
if(matrix[mid][c-1] < target) s = mid+1;
else e = mid-1;
}
// after the above binary search the s will store the row number of the target ,
// note that is after this binary search s=r (if target is greater than largest element matrix[r-1][c-1]
// and s=0 (s is in first row or less than the least element matrix[0][0]).
// since i have considered both these cases earlier using two if conditions , otherwise i can use
// this if condition here   **  if(s == r) return false; **
int left = 0, right = c-1;
while(left <= right){
mid = left + (right-left)/2;
if(matrix[s][mid] == target) return true;
else if(matrix[s][mid] > target) right = mid-1;
else left = mid+1;
}
// target is not present in matrix
return false;
}``````

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