Longest Increasing Path - Amazon Top Interview Questions


Problem Statement :


Given a two-dimensional integer matrix, find the length of the longest strictly increasing path. You can move up, down, left, or right.

Constraints

n, m ≤ 500 where n and m are the number of rows and columns in matrix

Example 1

Input

matrix = [
    [1, 3, 5],
    [0, 4, 6],
    [2, 2, 9]
]

Output

6

Explanation

The longest path is [0, 1, 3, 5, 6, 9]


Solution :



title-img 




                        Solution in C++ :

const int dir[4][2] = {{-1, 0}, {0, -1}, {0, 1}, {1, 0}};
int dfs(const vector<vector<int>> &matrix, vector<vector<int>> &dp, const int &r, const int &c) {
    if (dp[r][c] != -1) return dp[r][c];
    dp[r][c] = 1;
    for (int i = 0; i < 4; i++) {
        int nextR = r + dir[i][0];
        int nextC = c + dir[i][1];
        if (nextR < 0 || nextC < 0 || nextR >= matrix.size() || nextC >= matrix[0].size() ||
            matrix[nextR][nextC] <= matrix[r][c]) {
            continue;
        }
        dp[r][c] = max(dp[r][c], 1 + dfs(matrix, dp, nextR, nextC));
    }
    return dp[r][c];
}
int solve(vector<vector<int>> &matrix) {
    int m = matrix.size();
    if (m < 1) return 0;
    int n = matrix[0].size();
    vector<vector<int>> dp(m, vector<int>(n, -1));
    int ans = 0;
    for (int i = 0; i < m; i++) {
        for (int j = 0; j < n; j++) {
            if (dp[i][j] == -1) dfs(matrix, dp, i, j);
            ans = max(ans, dp[i][j]);
        }
    }
    return ans;
}
                    


                        Solution in Java :

import java.util.*;

class Solution {
    private int row;
    private int col;
    private int[][] dp;
    private boolean[][] visited;
    private int maxLen = 1;
    public int solve(int[][] matrix) {
        if (matrix == null || matrix.length == 0)
            return 0;
        row = matrix.length;
        col = matrix[0].length;
        dp = new int[row][col];

        for (int[] row : dp) Arrays.fill(row, -1);

        for (int i = 0; i < row; i++) {
            for (int j = 0; j < col; j++) {
                maxLen = Math.max(maxLen, dfs(i, j, matrix, -1));
            }
        }
        return maxLen;
    }

    private int dfs(int i, int j, int[][] matrix, int len) {
        if (i < 0 || j < 0 || i >= row || j >= col)
            return 0;

        if (matrix[i][j] <= len)
            return 0;

        if (dp[i][j] != -1)
            return dp[i][j];

        int l1 = dfs(i, j + 1, matrix, matrix[i][j]);
        int l2 = dfs(i, j - 1, matrix, matrix[i][j]);
        int l3 = dfs(i - 1, j, matrix, matrix[i][j]);
        int l4 = dfs(i + 1, j, matrix, matrix[i][j]);

        return dp[i][j] = 1 + Math.max(Math.max(l1, l2), Math.max(l3, l4));
    }
}
                    


                        Solution in Python : 
                            
class Solution:
    def solve(self, a):
        if not a or not a[0]:
            return 0
        n, m = len(a), len(a[0])
        dp = [[1 for j in range(m)] for i in range(n)]
        for _, i, j in sorted((a[i][j], i, j) for i in range(n) for j in range(m)):
            for ni, nj in ((i - 1, j), (i + 1, j), (i, j - 1), (i, j + 1)):
                if 0 <= ni < n and 0 <= nj < m and a[i][j] < a[ni][nj]:
                    dp[ni][nj] = max(dp[ni][nj], dp[i][j] + 1)
        return max(map(max, dp))


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