0579MediumArrayBinary SearchMatrix

Any Summit in a Height Field

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Trains the technique from

LeetCode 1901Find a Peak Element II

This is an original problem, written from a brief that listed the technique, the difficulty, the topics, the function shape and the input bounds — none of that problem's wording, examples, hints or editorials. The link is there so you can map your practice onto the standard set.

Same function shape, different story and different numbers.

A height field is stored in mat, where mat[r][c] is the altitude of cell r, c. No two cells that share an edge hold the same altitude.

A cell is a summit when its altitude is strictly above the altitude of every cell sharing an edge with it. Cells on the border have fewer neighbours to beat, and everything off the edge of the field counts as lower than any cell, so a summit always exists.

Return the position of a summit as the pair [r, c]. When the field holds more than one summit, any one of them is accepted.

Aim for a routine that does not look at every cell: a running time on the order of rows * log(columns) or columns * log(rows) is expected.

Examples

Example 1

Input: mat = [[1, 4, 3]]
Output: [0, 1]

Cell 0, 1 holds 4, which is above both 1 and 3, and it is the only summit in this field.

Example 2

Input: mat = [[5, 8, 6], [9, 20, 4]]
Output: [1, 1]

Cell 1, 1 holds 20 and beats 8 above it, 9 to its left and 4 to its right, so it is a summit.

Example 3

Input: mat = [[10, 2, 9], [3, 1, 4], [8, 5, 7]]
Output: [0, 0]

Cell 0, 0 holds 10 and beats 2 to its right and 3 below it. The cells at 0, 2 and at 2, 0 and at 2, 2 are summits too, so any of those four answers is accepted.

Constraints

m == mat.length
n == mat[i].length
1 <= m, n <= 500
1 <= mat[i][j] <= 10^5
No two cells sharing an edge hold equal altitudes.

More than one answer is valid — return any one of them.

The signature

The editor is preloaded with this. It matches the parent problem's shape, so a solution that works here transfers to a judge unchanged.

Python

def find_peak_grid(mat: list[list[int]]) -> list[int]:

Java

public int[] findPeakGrid(int[][] mat)

The intended approach

Time O(rows * log(columns)) — one scan of a column per halving.Space O(1) — only indices, plus the two-element answer.

Keep a block of columns that is known to contain a summit, starting with all of them. Scan the middle column of the block and take its tallest cell, which beats whichever of its own column neighbours exist. Compare it with the cells immediately left and right: if it beats both, it is a summit. If not, move the block to the side holding the taller of those two, because the tallest cell of that side is at least as tall and can never be beaten across the column just left behind, so a summit is trapped there. Each round halves the number of columns and costs one column scan.

Pythonentry point: find_peak_grid

def find_peak_grid(mat):
    # O(rows * log(cols)) time, O(1) space: halve the range of columns. In the
    # middle column take the tallest cell; it already beats its own column
    # neighbours. If it also beats the cells left and right of it, it is a
    # summit. Otherwise the taller side must contain a summit, because the
    # tallest cell of that half-block cannot be beaten from the column just
    # crossed, so the search continues on that side alone.
    rows = len(mat)
    cols = len(mat[0])
    lo = 0
    hi = cols - 1
    while lo <= hi:
        mid = (lo + hi) // 2
        top = 0
        for r in range(rows):
            if mat[r][mid] > mat[top][mid]:
                top = r
        here = mat[top][mid]
        left = mat[top][mid - 1] if mid > 0 else 0
        right = mat[top][mid + 1] if mid + 1 < cols else 0
        if here > left and here > right:
            return [top, mid]
        if left > here:
            hi = mid - 1
        else:
            lo = mid + 1
    return [-1, -1]

The obvious version

Visit every cell in row-major order and check it against each neighbour it has. Return the first cell that beats all of them.

This is not filler. It was written independently of the fast solution and the two were run against thousands of random inputs to confirm they agree — which is how the answer above is known to be right. Read it when the intended approach feels like a rabbit pulled out of a hat.

Python

def bf_find_peak_grid(mat):
    rows = len(mat)
    cols = len(mat[0])
    for r in range(rows):
        for c in range(cols):
            value = mat[r][c]
            if r > 0 and mat[r - 1][c] > value:
                continue
            if r + 1 < rows and mat[r + 1][c] > value:
                continue
            if c > 0 and mat[r][c - 1] > value:
                continue
            if c + 1 < cols and mat[r][c + 1] > value:
                continue
            return [r, c]
    return [-1, -1]

How this problem was checked

Reference solution executed against 11 test cases — 11 passed
3 worked examples confirmed by running the code
0 eight-word sequences shared with the parent problem's text
Agrees with an independent brute-force implementation on random inputs
Test set proven to catch deliberately broken solutions
Reviewed by a separate quality pass
Function arity matches the parent problem's

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