# 329. Longest Increasing Path in a Matrix

## LeetCode [329. Longest Increasing Path in a Matrix](https://github.com/txfs19260817/LeetCode-Go/blob/master/solutions/title/README.md)

### Description

Given an `m x n` integers `matrix`, return *the length of the longest increasing path in*`matrix`.

From each cell, you can either move in four directions: left, right, up, or down. You **may not** move **diagonally** or move **outside the boundary** (i.e., wrap-around is not allowed).

**Example 1:**

![](/files/RTD8nFsP87HO3i5KpkHe)

```
Input: matrix = [[9,9,4],[6,6,8],[2,1,1]]
Output: 4
Explanation: The longest increasing path is [1, 2, 6, 9].
```

**Constraints:**

* `m == matrix.length`
* `n == matrix[i].length`
* `1 <= m, n <= 200`
* `0 <= matrix[i][j] <= 2^31 - 1`

### Tags

Depth-first Search, Topological Sort, Memoization

### Solution

We build a 2D array `dp` with the same size of `matrix` to store the Longest Increasing Path starting from `matrix[i][j]`, with all elements initialized with 1. We also need an 1D array of all elements in descending order. It allows us to find the longest path in decreasing way. Then we perform BFS for each elements of this 1D array, and update the longest path.

### Complexity

* Time complexity: $$O(mn)$$
* Space complexity: $$O(mn)$$

### Code

```go
func longestIncreasingPath(matrix [][]int) int {
	ans, elements := 1, make([][]int, 0, len(matrix)*len(matrix[0]))
	for i, line := range matrix {
		for j, v := range line {
			elements = append(elements, []int{v, i, j})
		}
	}
	sort.Slice(elements, func(i, j int) bool {
		return elements[i][0] > elements[j][0]
	})
	dp, dirs := make([][]int, len(matrix)), [][]int{{1, 0}, {-1, 0}, {0, 1}, {0, -1}}
	for i := range dp {
		dp[i] = make([]int, len(matrix[0]))
		for j := range dp[i] {
			dp[i][j] = 1
		}
	}
	for _, e := range elements {
		v, i, j := e[0], e[1], e[2]
		for _, d := range dirs {
			if x, y := i+d[0], j+d[1]; x >= 0 && y >= 0 && x < len(matrix) && y < len(matrix[0]) && matrix[x][y] > v && dp[i][j] < dp[x][y]+1 {
				dp[i][j] = dp[x][y] + 1
			}
		}
		if dp[i][j] > ans {
			ans = dp[i][j]
		}
	}
	return ans
}
```

## Reference

1. [329. Longest Increasing Path in a Matrix](https://zhuanlan.zhihu.com/p/83101599)


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