If I was at an interview, and wrote this code, what would you think? Please be brutal.
Time it took to wrote it: 13 minutes
Problem:
Write an efficient algorithm that searches for a value in an m x n matrix. This matrix has the following properties:
Integers in each row are sorted from left to right. The first integer of each row is greater than the last integer of the previous row. For example,
Consider the following matrix:
[ [1, 3, 5, 7], [10, 11, 16, 20], [23, 30, 34, 50] ]
Given target = 3, return true.
public boolean searchMatrix(int[][] matrix, int target) {
int lastCol = matrix[0].length-1;
int row=0;
while(row!=matrix.length && lastCol>=0){
if(matrix[row][lastCol]==target){
return true;
}
if(matrix[row][lastCol]>target){
lastCol--;
if(binarySearch(matrix[row], 0, lastCol, target)!=-1){
return true;
}
}
row++;
}
return false;
}
public int binarySearch(int[] arr, int low, int high, int target){
while(low<=high){
int mid = (low + high)/2;
if(arr[mid]==target){
return mid;
}
else if(arr[mid]<target){
low = mid+1;
}
else if(arr[mid]>target){
high = mid-1;
}
}
return -1;
}
Space complexity = \$\mathcal{O}(1)\$
Time Complexity = \$\mathcal{O}(\log(n!))\$
row = mid/ arr[0].length()
andcol = mid % arr[0].length()
. If no-one else does, I'll post an answer tomorrow, but I need to get some sleep. \$\endgroup\$