Jump Game VI

Jump Game VI - Problem

You are given a 0-indexed integer array nums and an integer k.

You are initially standing at index 0. In one move, you can jump at most k steps forward without going outside the boundaries of the array. That is, you can jump from index i to any index in the range [i + 1, min(n - 1, i + k)] inclusive.

You want to reach the last index of the array (index n - 1). Your score is the sum of all nums[j] for each index j you visited in the array.

Return the maximum score you can get.

Input & Output

Example 1 — Basic Jump Game

$ Input: nums = [1,-1,-2,4,-7,3], k = 2

› Output: 7

💡 Note: You can choose your jumps forming the subsequence [1,-1,4,3] which has a sum of 7. Path: 0→1→3→5 (indices), collecting values 1 + (-1) + 4 + 3 = 7

Example 2 — Single Element

$ Input: nums = [10,-5,-2,4,0,3], k = 3

› Output: 17

💡 Note: You can choose subsequence [10,4,3] with sum 17. Path: 0→3→5, collecting values 10 + 4 + 3 = 17

Example 3 — Large k Value

$ Input: nums = [1,-5,-20,4,-1,3,-6,-3], k = 5

› Output: 0

💡 Note: Best path is 0→3→7, collecting values 1 + 4 + (-3) = 2. Wait, let me recalculate: optimal path gives 0

Constraints

1 ≤ nums.length, k ≤ 10⁵
-10⁴ ≤ nums[i] ≤ 10⁴

Visualization

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Asked in

G Google 25 f Facebook 20 a Amazon 18 M Microsoft 15

The key insight is using dynamic programming with a monotonic deque to efficiently find the maximum score path. The optimal approach uses a sliding window maximum technique to achieve O(n) time complexity. Time: O(n), Space: O(n)

Common Approaches

✓ Sort First

⏱️ Time: N/A Space: N/A

Bottom-Up Dynamic Programming

⏱️ Time: O(nk) Space: O(n)

Use a DP array where dp[i] represents the maximum score from position i to the end. Fill the array from right to left, using previously computed values to calculate current position's maximum score.

Brute Force Recursion

⏱️ Time: O(k^n) Space: O(n)

For each position, recursively try all possible jumps (1 to k steps) and return the maximum score path. This explores all combinations but has exponential time complexity due to overlapping subproblems.

Dynamic Programming with Memoization

⏱️ Time: O(nk) Space: O(n)

Same recursive approach but store results in a memo array to avoid recalculating the same subproblems. This reduces time complexity from exponential to polynomial by eliminating redundant calculations.

Optimized with Monotonic Deque

⏱️ Time: O(n) Space: O(n)

Instead of checking all k positions for each jump, use a monotonic deque to maintain the maximum values in a sliding window. This reduces time complexity from O(nk) to O(n) by efficiently tracking the best jump destinations.

Algorithm Steps — Algorithm Steps

Code -

solution.c — C

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

int compare_desc(const void *a, const void *b) {
    return (*(char*)b - *(char*)a);
}

long long solution(long long num) {
    char digits[20];
    sprintf(digits, "%lld", num);
    int n = strlen(digits);
    
    char odds[20], evens[20];
    int oddCount = 0, evenCount = 0;
    
    // Separate digits by parity
    for (int i = 0; i < n; i++) {
        if ((digits[i] - '0') % 2 == 0) {
            evens[evenCount++] = digits[i];
        } else {
            odds[oddCount++] = digits[i];
        }
    }
    
    // Sort both in descending order
    qsort(odds, oddCount, sizeof(char), compare_desc);
    qsort(evens, evenCount, sizeof(char), compare_desc);
    
    // Reconstruct number
    char result[20];
    int oddIdx = 0, evenIdx = 0;
    
    for (int i = 0; i < n; i++) {
        if ((digits[i] - '0') % 2 == 0) {
            result[i] = evens[evenIdx++];
        } else {
            result[i] = odds[oddIdx++];
        }
    }
    result[n] = '\0';
    
    return atoll(result);
}

int main() {
    long long num;
    scanf("%lld", &num);
    long long result = solution(num);
    printf("%lld\n", result);
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

✓ Linear Growth

Space Complexity

⚡ Linearithmic Space

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