Length of the Longest Increasing Path

Length of the Longest Increasing Path - Problem

You are given a 2D array of integers coordinates of length n and an integer k, where 0 <= k < n. coordinates[i] = [x_i, y_i] indicates the point (x_i, y_i) in a 2D plane.

An increasing path of length m is defined as a list of points (x_1, y_1), (x_2, y_2), (x_3, y_3), ..., (x_m, y_m) such that:

x_i < x_{i+1} and y_i < y_{i+1} for all i where 1 <= i < m
(x_i, y_i) is in the given coordinates for all i where 1 <= i <= m

Return the maximum length of an increasing path that contains coordinates[k].

Input & Output

Example 1 — Basic Increasing Path

$ Input: coordinates = [[1,1],[2,3],[4,2],[5,4]], k = 0

› Output: 3

💡 Note: Starting from coordinates[0] = (1,1), we can form path (1,1) → (2,3) → (5,4) with length 3

Example 2 — Middle Point

$ Input: coordinates = [[1,1],[2,3],[4,2],[5,4]], k = 1

› Output: 2

💡 Note: Starting from coordinates[1] = (2,3), we can form path (2,3) → (5,4) with length 2

Example 3 — Single Point

$ Input: coordinates = [[3,4]], k = 0

› Output: 1

💡 Note: Only one point exists, so maximum path length is 1

Constraints

1 ≤ coordinates.length ≤ 10³
0 ≤ k < coordinates.length
coordinates[i].length == 2
-10⁴ ≤ coordinates[i][0], coordinates[i][1] ≤ 10⁴

Visualization

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Asked in

G Google 25 M Microsoft 18 a Amazon 15

The key insight is to use dynamic programming with memoization to find the longest increasing path starting from each point. We recursively explore all valid next points where both x and y coordinates are strictly greater. Best approach is DP with memoization. Time: O(n²), Space: O(n)

Common Approaches

✓ Sliding Window

⏱️ Time: N/A Space: N/A

Brute Force DFS

⏱️ Time: O(2^n) Space: O(n)

For each point, use DFS to explore all valid increasing paths and find the maximum length path that includes the target point at index k.

Dynamic Programming

⏱️ Time: O(n²) Space: O(n)

For each point, calculate the maximum length of increasing path starting from that point using dynamic programming with memoization.

Sort + Binary Search Optimization

⏱️ Time: O(n²) Space: O(n)

Sort all points by x-coordinate, then for each point, we only need to check points that come after it in sorted order, using binary search to find valid y-coordinates.

Algorithm Steps — Algorithm Steps

Code -

solution.c — C

#include <stdio.h>
#include <string.h>
#include <stdlib.h>

static int dp[1000];
static int coordinates[1000][2];
static int n;

int getLongestPath(int i) {
    if (dp[i] != -1) {
        return dp[i];
    }
    
    dp[i] = 1;
    int x = coordinates[i][0];
    int y = coordinates[i][1];
    
    for (int j = 0; j < n; j++) {
        if (i != j) {
            int nx = coordinates[j][0];
            int ny = coordinates[j][1];
            if (nx > x && ny > y) {
                int pathLen = 1 + getLongestPath(j);
                if (pathLen > dp[i]) {
                    dp[i] = pathLen;
                }
            }
        }
    }
    
    return dp[i];
}

int solution(int k) {
    for (int i = 0; i < n; i++) {
        dp[i] = -1;
    }
    
    return getLongestPath(k);
}

int main() {
    char coordsStr[10000];
    int k;
    
    fgets(coordsStr, sizeof(coordsStr), stdin);
    coordsStr[strcspn(coordsStr, "\n")] = 0;
    scanf("%d", &k);
    
    // Parse coordinates
    n = 0;
    char* ptr = coordsStr + 1; // Skip first '['
    
    while (*ptr && *ptr != ']') {
        if (*ptr == '[') {
            ptr++;
            coordinates[n][0] = strtol(ptr, &ptr, 10);
            ptr++; // Skip comma
            coordinates[n][1] = strtol(ptr, &ptr, 10);
            ptr++; // Skip ']'
            n++;
        }
        ptr++;
    }
    
    int result = solution(k);
    printf("%d\n", result);
    
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

⚠ Quadratic Growth

Space Complexity

⚡ Linearithmic Space

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