LeetCode – Longest Increasing Path in a Matrix (Java)

Given an integer matrix, get the length of the longest increasing path.

https://leetcode.com/problems/longest-increasing-path-in-a-matrix/

Java Solution 1 - Naive DFS

public int longestIncreasingPath(int[][] matrix) {
    int[] max = new int[1];
    for (int i = 0; i < matrix.length; i++) {
        for (int j = 0; j < matrix[0].length; j++) {
            dfs(matrix, i, j, max, 1);
        }
    }
 
    return max[0];
}
 
public void dfs(int[][] matrix, int i, int j, int[] max, int len) {
    max[0] = Math.max(max[0], len);
 
    int m = matrix.length;
    int n = matrix[0].length;
 
    int[] dx = {-1, 0, 1, 0};
    int[] dy = {0, 1, 0, -1};
 
    for (int k = 0; k < 4; k++) {
        int x = i + dx[k];
        int y = j + dy[k];
 
        if (x >= 0 && x < m && y >= 0 && y < n && matrix[x][y] > matrix[i][j]) {
            dfs(matrix, x, y, max, len + 1);
        }
    }
}

This naive DFS solution's time complexity is O(m*n*4^(m+n)).

Java Solution 2- Optimization with memorization

A common approach to improve DFS is through memorization.

public int longestIncreasingPath(int[][] matrix) {
    int result = 0;
    int m = matrix.length;
    int n = matrix[0].length;
 
    int[][] mem = new int[m][n];
    for (int i = 0; i < m; i++) {
        for (int j = 0; j < n; j++) {
            int t = helper(matrix, mem, i, j);
            result = Math.max(result, t);
        }
    }
 
    return result;
}
 
private int helper(int[][] matrix, int[][] mem, int i, int j) {
    if (mem[i][j] > 0) {
        return mem[i][j];
    }
 
    int[] dx = {-1, 0, 1, 0};
    int[] dy = {0, 1, 0, -1};
 
    for (int k = 0; k < 4; k++) {
        int x = i + dx[k];
        int y = j + dy[k];
 
        if (x >= 0 && y >= 0
                && x < matrix.length
                && y < matrix[0].length
                && matrix[x][y] > matrix[i][j]) {
            mem[i][j] = Math.max(mem[i][j], helper(matrix, mem, x, y));
        }
    }
 
    return ++mem[i][j];
}

Because of the memorization matrix, the upper bound time complexity of the DFS is O(m*n). With the loop in the main method, the overall time complexity is O(m^2 * n^2)

The following shows the first 4 elements' DFS during iteration.

Category >> Algorithms  
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  • Rohith Uppala

    Nope. The greater condition will take care of that.

  • Eldo Joseph

    Shouldn’t we track the visited coordinates?

  • Apurv Dubey

    that’s awesome solutions!