Odd Even Jump

Odd Even Jump - Problem

Array Dynamic Programming Stack Sorting Monotonic Stack Ordered Set Hard

You are given an integer array arr. From some starting index, you can make a series of jumps. The (1st, 3rd, 5th, ...) jumps in the series are called odd-numbered jumps, and the (2nd, 4th, 6th, ...) jumps in the series are called even-numbered jumps. Note that the jumps are numbered, not the indices.

You may jump forward from index i to index j (with i < j) in the following way:

During odd-numbered jumps (i.e., jumps 1, 3, 5, ...), you jump to the index j such that arr[i] <= arr[j] and arr[j] is the smallest possible value. If there are multiple such indices j, you can only jump to the smallest such index j.
During even-numbered jumps (i.e., jumps 2, 4, 6, ...), you jump to the index j such that arr[i] >= arr[j] and arr[j] is the largest possible value. If there are multiple such indices j, you can only jump to the smallest such index j.

It may be the case that for some index i, there are no legal jumps.

A starting index is good if, starting from that index, you can reach the end of the array (index arr.length - 1) by jumping some number of times (possibly 0 or more than once). Return the number of good starting indices.

Input & Output

Example 1 — Basic Jump Sequence

$ Input: arr = [10,13,12,14,15]

› Output: 2

💡 Note: From index 0: 10→13(odd)→12(even)→14(odd)→15 reaches end. From index 2: 12→14(odd)→15 reaches end. Positions 1,3,4 cannot reach end due to jump constraints.

Example 2 — Single Element

$ Input: arr = [2]

› Output: 1

💡 Note: Single element array - position 0 is already at the end, so it's a good starting index.

Example 3 — No Valid Jumps

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

› Output: 3

💡 Note: From index 0: 5→no valid odd jump (need ≥5). From index 2: 3→4(odd) reaches end. From index 3: 4→no valid jump. Only indices 2, 3, 4 are good.

Constraints

1 ≤ arr.length ≤ 2 × 10⁴
0 ≤ arr[i] < 10⁵

Visualization

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Asked in

G Google 42 f Facebook 23 a Amazon 18

The key insight is to work backwards using dynamic programming after building next jump maps with monotonic stacks. We first find the next valid position for each jump type, then determine reachability from the end. Best approach uses monotonic stack + DP: Time: O(n log n), Space: O(n)

Common Approaches

✓ Brute Force - DFS from Each Position

⏱️ Time: O(n × 2^n) Space: O(n)

For each starting position, use DFS to explore all possible jump sequences alternating between odd and even jumps. Check if any sequence reaches the last index.

Dynamic Programming with Monotonic Stack

⏱️ Time: O(n log n) Space: O(n)

Build next jump arrays using monotonic stacks, then use DP working backwards to determine which positions can reach the end with odd or even jump counts.

Brute Force - DFS from Each Position — Algorithm Steps

Step 1: For each starting index, begin DFS with odd jump
Step 2: At each position, find all valid next positions based on jump type
Step 3: Recursively explore each valid next position with alternating jump type
Step 4: Count positions that can reach the end

Visualization

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Step-by-Step Walkthrough

Start DFS

Begin DFS from each index with odd jump

Find Valid Jumps

Check all forward positions for valid next jumps

Recursive Exploration

Recursively explore each valid path alternating jump types

Code -

solution.c — C

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

bool dfs(int* arr, int n, int pos, bool isOddJump) {
    if (pos == n - 1) {
        return true;
    }
    
    int bestPos = -1;
    if (isOddJump) {
        int bestVal = INT_MAX;
        for (int nextPos = pos + 1; nextPos < n; nextPos++) {
            if (arr[pos] <= arr[nextPos]) {
                if (arr[nextPos] < bestVal || (arr[nextPos] == bestVal && nextPos < bestPos)) {
                    bestVal = arr[nextPos];
                    bestPos = nextPos;
                }
            }
        }
    } else {
        int bestVal = INT_MIN;
        for (int nextPos = pos + 1; nextPos < n; nextPos++) {
            if (arr[pos] >= arr[nextPos]) {
                if (arr[nextPos] > bestVal || (arr[nextPos] == bestVal && nextPos < bestPos)) {
                    bestVal = arr[nextPos];
                    bestPos = nextPos;
                }
            }
        }
    }
    
    if (bestPos != -1) {
        return dfs(arr, n, bestPos, !isOddJump);
    }
    return false;
}

int solution(int* arr, int n) {
    int count = 0;
    
    for (int i = 0; i < n; i++) {
        if (dfs(arr, n, i, true)) {
            count++;
        }
    }
    
    return count;
}

void parseArray(const char* str, int* arr, int* size) {
    *size = 0;
    const char* p = str;
    while (*p && *p != '[') p++;
    if (*p == '[') p++;
    while (*p && *p != ']') {
        while (*p == ' ' || *p == ',') p++;
        if (*p == ']' || *p == '\0') break;
        arr[(*size)++] = (int)strtol(p, (char**)&p, 10);
    }
}

int main() {
    char line[10000];
    fgets(line, sizeof(line), stdin);
    
    // Parse JSON array
    int arr[1000];
    int n = 0;
    char* token = strtok(line + 1, ",]");
    while (token != NULL) {
        arr[n++] = atoi(token);
        token = strtok(NULL, ",]");
    }
    
    int result = solution(arr, n);
    printf("%d\n", result);
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n × 2^n)

For each of n starting positions, we may explore up to 2^n possible jump sequences

⚠ Quadratic Growth

Space Complexity

O(n)

Recursion stack depth can go up to n levels

⚡ Linearithmic Space

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Constraints

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Related Problems

Common Approaches

Brute Force - DFS from Each Position — Algorithm Steps

Visualization

Code -

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