First I wrote a bruteforce solution, but it didn't pass because of too many calls, the time complexity is O(m*n):

    public int leftMostColumnWithOne(BinaryMatrix binaryMatrix) {
        List<Integer> dimensions = binaryMatrix.dimensions();
        int m = dimensions.get(0), n = dimensions.get(1);
        for(int j=0;j<n;j++){
            for(int i=0;i<m;i++){
                if(binaryMatrix.get(i,j)==1)
                    return j;
            }
        }
        return -1;
    }

The I improve the solution above with binary search, as the rows are sorted, so we can use binary search to find the column, the time complexity is O(m*logn).

    public int leftMostColumnWithOne(BinaryMatrix binaryMatrix) {
        List<Integer> dimensions = binaryMatrix.dimensions();
        int m = dimensions.get(0), n = dimensions.get(1);
        int l = 0, r = n, mid = 0;
        while(l<r){
            mid = (l+r)/2;
            if(hasOne(binaryMatrix, mid, m))
                r = mid;
            else
                l = mid+1;
        }
        return r==n?-1: r; //if there are no columns contain 1, the r would be n at last.
    }
    
    private boolean hasOne(BinaryMatrix binaryMatrix, int col, int m){
        for(int i=0;i<m;i++){
            if(binaryMatrix.get(i, col)==1)
                return true;
        }
        return false;
    }