Find Peak Element II

Description:

There is an integer matrix which has the following features:

• The numbers in adjacent positions are different.
• The matrix has n rows and m columns.
• For all i < m, A[0][i] < A[1][i] && A[n - 2][i] > A[n - 1][i].
• For all j < n, A[j][0] < A[j][1] && A[j][m - 2] > A[j][m - 1].

We define a position P is a peek if:

A[j][i] > A[j+1][i] && A[j][i] > A[j-1][i] && A[j][i] > A[j][i+1] && A[j][i] > A[j][i-1]

Find a peak element in this matrix. Return the index of the peak.

Example

Given a matrix:

[
  [1 ,2 ,3 ,6 ,5],
  [16,41,23,22,6],
  [15,17,24,21,7],
  [14,18,19,20,10],
  [13,14,11,10,9]
]

return index of 41 (which is [1,1]) or index of 24 (which is [2,2])

Note

The matrix may contains multiple peeks, find any of them.

Challenge

Solve it in O(n+m) time.

If you come up with an algorithm that you thought it is O(n log m) or O(m log n), can you prove it is actually O(n+m) or propose a similar but O(n+m) algorithm?

Solution:

Reference: http://courses.csail.mit.edu/6.006/spring11/lectures/lec02.pdf

class Solution {
public:
/**
* @param A: An integer matrix
* @return: The index of the peak
*/
vector findPeakII(vector<vector > A) {
auto m = (int)A.size();
auto n = (int)A[0].size();
int up = 0;
int down = m-1;
int left = 0;
int right = n-1;
while (left <= right && up <= down) {
int midRow = up+down>>1;
int maxCol = findPeakInRow(A, midRow, left, right);
if (A[midRow][maxCol] < A[midRow-1][maxCol])
down = midRow-1;
else if (A[midRow][maxCol] < A[midRow+1][maxCol])
up = midRow+1;
else return vector({midRow, maxCol});

int midCol = left+right>>1;
int maxRow = findPeakInCol(A, midCol, up, down);
if (A[maxRow][midCol] < A[maxRow][midCol-1])
right = midCol-1;
else if (A[maxRow][midCol] < A[maxRow][midCol+1])
left = midCol+1;
else return vector({maxRow, midCol});
}
return vector({up, left});
}

private:
int findPeakInRow(vector<vector>& A, int i, int left, int right) {
int rc = left;
for (int j = left+1; j < right; ++j)
if (A[i][j] > A[i][rc])
rc = j;
return rc;
}

int findPeakInCol(vector<vector>& A, int j, int up, int down) {
int rc = up;
for (int i = up+1; i < down; ++i)
if (A[i][j] > A[rc][j])
rc = i;
return rc;
}
};