1901. Find a Peak Element II

Description

A peak element in a 2D grid is an element that is strictly greater than all of its adjacent neighbors to the left, right, top, and bottom.

Given a 0-indexed m x n matrix mat where no two adjacent cells are equal, find any peak element mat[i][j] and return the length 2 array [i,j].

You may assume that the entire matrix is surrounded by an outer perimeter with the value -1 in each cell.

You must write an algorithm that runs in O(m log(n)) or O(n log(m)) time.

Example 1:

Input: mat = [[1,4],[3,2]]
Output: [0,1]
Explanation: Both 3 and 4 are peak elements so [1,0] and [0,1] are both acceptable answers.


Example 2:

Input: mat = [[10,20,15],[21,30,14],[7,16,32]]
Output: [1,1]
Explanation: Both 30 and 32 are peak elements so [1,1] and [2,2] are both acceptable answers.


Constraints:

• m == mat.length
• n == mat[i].length
• 1 <= m, n <= 500
• 1 <= mat[i][j] <= 105
• No two adjacent cells are equal.

Solutions

Solution 1: Binary Search

Let $m$ and $n$ be the number of rows and columns of the matrix, respectively.

The problem asks us to find a peak, and the time complexity should be $O(m \times \log n)$ or $O(n \times \log m)$. Therefore, we can consider using binary search.

We consider the maximum value of the $i$-th row, and denote its index as $j$.

If $mat[i][j] > mat[i + 1][j]$, then there must be a peak in the rows $[0,..i]$. We only need to find the maximum value in these rows. Similarly, if $mat[i][j] < mat[i + 1][j]$, then there must be a peak in the rows $[i + 1,..m - 1]$. We only need to find the maximum value in these rows.

Why is the above method correct? We can prove it by contradiction.

If $mat[i][j] > mat[i + 1][j]$, suppose there is no peak in the rows $[0,..i]$. Then $mat[i][j]$ is not a peak. Since $mat[i][j]$ is the maximum value of the $i$-th row, and $mat[i][j] > mat[i + 1][j]$, then $mat[i][j] < mat[i - 1][j]$. We continue to consider from the $(i - 1)$-th row upwards, and the maximum value of each row is less than the maximum value of the previous row. When we traverse to $i = 0$, since all elements in the matrix are positive integers, and the values of the cells around the matrix are $-1$. Therefore, at the 0-th row, its maximum value is greater than all its adjacent elements, so the maximum value of the 0-th row is a peak, which contradicts the assumption. Therefore, there must be a peak in the rows $[0,..i]$.

For the case where $mat[i][j] < mat[i + 1][j]$, we can prove in a similar way that there must be a peak in the rows $[i + 1,..m - 1]$.

Therefore, we can use binary search to find the peak.

We perform binary search on the rows of the matrix, initially with the search boundaries $l = 0$, $r = m - 1$. Each time, we find the middle row $mid$ and find the index $j$ of the maximum value of this row. If $mat[mid][j] > mat[mid + 1][j]$, then we search for the peak in the rows $[0,..mid]$, i.e., update $r = mid$. Otherwise, we search for the peak in the rows $[mid + 1,..m - 1]$, i.e., update $l = mid + 1$. When $l = r$, we find the position $[l, j_l]$ of the peak, where $j_l$ is the index of the maximum value of the $l$-th row.

The time complexity is $O(n \times \log m)$, where $m$ and $n$ are the number of rows and columns of the matrix, respectively. The time complexity of binary search is $O(\log m)$, and each time we perform binary search, we need to traverse all elements of the $mid$-th row, with a time complexity of $O(n)$. The space complexity is $O(1)$.

• class Solution {
public int[] findPeakGrid(int[][] mat) {
int l = 0, r = mat.length - 1;
int n = mat[0].length;
while (l < r) {
int mid = (l + r) >> 1;
int j = maxPos(mat[mid]);
if (mat[mid][j] > mat[mid + 1][j]) {
r = mid;
} else {
l = mid + 1;
}
}
return new int[] {l, maxPos(mat[l])};
}

private int maxPos(int[] arr) {
int j = 0;
for (int i = 1; i < arr.length; ++i) {
if (arr[j] < arr[i]) {
j = i;
}
}
return j;
}
}

• class Solution {
public:
vector<int> findPeakGrid(vector<vector<int>>& mat) {
int l = 0, r = mat.size() - 1;
while (l < r) {
int mid = (l + r) >> 1;
int j = distance(mat[mid].begin(), max_element(mat[mid].begin(), mat[mid].end()));
if (mat[mid][j] > mat[mid + 1][j]) {
r = mid;
} else {
l = mid + 1;
}
}
int j = distance(mat[l].begin(), max_element(mat[l].begin(), mat[l].end()));
return {l, j};
}
};

• class Solution:
def findPeakGrid(self, mat: List[List[int]]) -> List[int]:
l, r = 0, len(mat) - 1
while l < r:
mid = (l + r) >> 1
j = mat[mid].index(max(mat[mid]))
if mat[mid][j] > mat[mid + 1][j]:
r = mid
else:
l = mid + 1
return [l, mat[l].index(max(mat[l]))]


• func findPeakGrid(mat [][]int) []int {
maxPos := func(arr []int) int {
j := 0
for i := 1; i < len(arr); i++ {
if arr[i] > arr[j] {
j = i
}
}
return j
}
l, r := 0, len(mat)-1
for l < r {
mid := (l + r) >> 1
j := maxPos(mat[mid])
if mat[mid][j] > mat[mid+1][j] {
r = mid
} else {
l = mid + 1
}
}
return []int{l, maxPos(mat[l])}
}

• function findPeakGrid(mat: number[][]): number[] {
let [l, r] = [0, mat.length - 1];
while (l < r) {
const mid = (l + r) >> 1;
const j = mat[mid].indexOf(Math.max(...mat[mid]));
if (mat[mid][j] > mat[mid + 1][j]) {
r = mid;
} else {
l = mid + 1;
}
}
return [l, mat[l].indexOf(Math.max(...mat[l]))];
}


• impl Solution {
pub fn find_peak_grid(mat: Vec<Vec<i32>>) -> Vec<i32> {
let mut l: usize = 0;
let mut r: usize = mat.len() - 1;
while l < r {
let mid: usize = (l + r) >> 1;
let j: usize = mat[mid]
.iter()
.position(|&x| x == *mat[mid].iter().max().unwrap())
.unwrap();
if mat[mid][j] > mat[mid + 1][j] {
r = mid;
} else {
l = mid + 1;
}
}
let j: usize = mat[l]
.iter()
.position(|&x| x == *mat[l].iter().max().unwrap())
.unwrap();
vec![l as i32, j as i32]
}
}