# My solution using Binary Search in C++

• class Solution
{
public:
int kthSmallest(vector<vector<int>>& matrix, int k)
{
int n = matrix.size();
int le = matrix[0][0], ri = matrix[n - 1][n - 1];
int mid = 0;
while (le < ri)
{
mid = le + (ri-le)/2;
int num = 0;
for (int i = 0; i < n; i++)
{
int pos = upper_bound(matrix[i].begin(), matrix[i].end(), mid) - matrix[i].begin();
num += pos;
}
if (num < k)
{
le = mid + 1;
}
else
{
ri = mid;
}
}
return le;
}
};

• @光速小子 what is the complexity of your solution? Your code could be cleaner and may need better variable names.

• Fails matrix=[[2000000000]], k=1. Better start with INT_MIN/INT_MAX or matrix[0][0]/matrix[n-1][n-1]. But nice solution.

• @StefanPochmann
Yes,you're right.
There are some parts of this code should be improved. Thank you very very much.

• @agave
The complexity of my solution is O(nlog(n)log(N)), n is the number of rows.
This is my first time to share my code and my variable name are so thoughtless.
Thank you for correcting me. I will do better next time.

• @光速小子 said in My solution using Binary Search in C++:

@agave
The complexity of my solution is O(rlogn), r is the number of rows.

Better don't redefine names. The problem statement already defined n to be the number of rows (and columns).

I will do better next time.

You can also edit your post and do better this time :-)

• @StefanPochmann
I didn't know I can edit it...Sorry for that.
I am editing it now.

• @StefanPochmann
I think this time the code is clearer.

• Easier to read C++ implementation according to your idea:

int kthSmallest(vector<vector<int>>& matrix, int k) {
int n = matrix.size();

int left = matrix[0][0];
int right = matrix[n-1][n-1];

while (left < right) {
int midv = (left + right) / 2;

int count = 0;  // number of elements no bigger than midv
for (int i = 0; i < n; i++) {
vector<int>& row = matrix[i];
count += std::upper_bound(row.begin(), row.end(), midv) - row.begin();
}

if (count < k) {
left = midv + 1;
} else {
right = midv;
}
}
return left;
}

• @storypku thank you.

• The time complexity is O(n * log(n) * log(N)), where N is the search space that ranges from the smallest element to the biggest element.

You can argue that int implies N = 2^32, so log(N) is constant. I guess this is where O(n * log(n)) comes from.

I thought this idea was weird for a while. Then I noticed the previous problem 377. Combination Sum IV is pretty much doing the same thing, so this idea may actually be intended. Nice one!

• Thanks for the solution. Java version. Thanks @StefanPochmann 's comment.

int n = matrix.length;

int left = matrix[0][0], right = matrix[n - 1][n - 1];

while (left < right) {
int mid = (left + right) / 2;
int count = 0; // number of elements no bigger than mid

for (int i = 0; i < n; i++) {
int[] row = matrix[i];

int t_left = 0, t_right = row.length;
while (t_left < t_right) {
int t_mid = (t_left + t_right) / 2;
int value = row[t_mid];
if(value > mid) {
t_right = t_mid;
} else {
t_left = t_mid + 1;
}
}
count += t_left;
}

if (count < k) {
left = mid + 1;
} else {
right = mid;
}
}
return left;

• @jianqing The inner part should be binary search, too. At least upper_bound is.

• This post is deleted!

• Actually, the inner part can be done in O(n) time.

class Solution {
public:
int kthSmallest(vector<vector<int>>& matrix, int k) {
int n=matrix.size();
int l=matrix[0][0], r=matrix[n-1][n-1], mid;
while(l<r){
mid=l+r>>1;
int cnt=0, j=n-1;
for(int i=0;i<n;i++){
while(j>=0&&matrix[i][j]>mid)
j--;
cnt+=j+1;
}
if(cnt<k)
l=mid+1;
else
r=mid;
}
return l;
}
};

• @storypku hi, thanks for your sharing.
I have a question:

• you return "left" in the last line. But how do u know the "left" is a value in the matrix?

• @csxuejin got the same question for this solution.

• Because the loop invariant is left<=Solution<=right. The moment it quits the loop, we also know another condition is true: left>=right.

left<=Solution<=right and left>=right means left==Solution==right.

• @o_sharp Clear now. Thanks:)

• I like this solution, but I'm a little confused. I always thought you couldn't do binary search on this sort of matrix - which was why we have to use saddleback searching. Can someone help correct my misunderstanding?

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