Odd Even Jump - Practice Coding Problems

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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Understanding the Visualization

Input Array

Array with jump constraints: odd jumps go to smallest ≥ value, even jumps to largest ≤ value

Jump Rules

Alternate between odd and even jumps, must jump forward

Count Good Starts

Count positions that can reach the last index

Key Takeaway

🎯 Key Insight: Use monotonic stacks to efficiently find next jump targets, then DP backwards to determine reachability

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

Approach	Time	Space	Notes
✓ Dynamic Programming with Monotonic Stack	O(n log n)	O(n)	Use monotonic stacks to find next jumps and DP to determine reachability backwards from end
Brute Force - DFS from Each Position	O(n × 2^n)	O(n)	Try all possible jump sequences from each starting index using DFS

Dynamic Programming with Monotonic Stack — Algorithm Steps

Step 1: Use monotonic stack to find next valid position for odd jumps
Step 2: Use monotonic stack to find next valid position for even jumps
Step 3: Work backwards with DP to mark positions reachable with odd/even jumps
Step 4: Count positions that can reach end starting with odd jump

Visualization

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Step-by-Step Walkthrough

Build Jump Maps

Use monotonic stacks to find next valid jump positions

DP Backwards

Work from end to start determining reachability

Count Results

Sum positions that can reach end with odd jump

Code -

solution.c — C

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

typedef struct {
    int index;
    int value;
} IndexValue;

int compareAsc(const void* a, const void* b) {
    IndexValue* ia = (IndexValue*)a;
    IndexValue* ib = (IndexValue*)b;
    if (ia->value != ib->value) {
        return ia->value - ib->value;
    }
    return ia->index - ib->index;
}

int compareDesc(const void* a, const void* b) {
    IndexValue* ia = (IndexValue*)a;
    IndexValue* ib = (IndexValue*)b;
    if (ia->value != ib->value) {
        return ib->value - ia->value;
    }
    return ia->index - ib->index;
}

int solution(int* arr, int n) {
    if (n <= 1) return n;
    
    int* nextHigher = (int*)malloc(n * sizeof(int));
    int* nextLower = (int*)malloc(n * sizeof(int));
    int* stack = (int*)malloc(n * sizeof(int));
    IndexValue* indices = (IndexValue*)malloc(n * sizeof(IndexValue));
    
    for (int i = 0; i < n; i++) {
        nextHigher[i] = -1;
        nextLower[i] = -1;
        indices[i].index = i;
        indices[i].value = arr[i];
    }
    
    // Find next higher for odd jumps
    qsort(indices, n, sizeof(IndexValue), compareAsc);
    
    int stackSize = 0;
    for (int i = 0; i < n; i++) {
        int idx = indices[i].index;
        while (stackSize > 0 && stack[stackSize-1] < idx) {
            nextHigher[stack[stackSize-1]] = idx;
            stackSize--;
        }
        stack[stackSize++] = idx;
    }
    
    // Find next lower for even jumps
    qsort(indices, n, sizeof(IndexValue), compareDesc);
    
    stackSize = 0;
    for (int i = 0; i < n; i++) {
        int idx = indices[i].index;
        while (stackSize > 0 && stack[stackSize-1] < idx) {
            nextLower[stack[stackSize-1]] = idx;
            stackSize--;
        }
        stack[stackSize++] = idx;
    }
    
    // DP
    bool* canReachOdd = (bool*)malloc(n * sizeof(bool));
    bool* canReachEven = (bool*)malloc(n * sizeof(bool));
    
    for (int i = 0; i < n; i++) {
        canReachOdd[i] = false;
        canReachEven[i] = false;
    }
    canReachOdd[n-1] = canReachEven[n-1] = true;
    
    for (int i = n - 2; i >= 0; i--) {
        if (nextHigher[i] != -1) {
            canReachOdd[i] = canReachEven[nextHigher[i]];
        }
        if (nextLower[i] != -1) {
            canReachEven[i] = canReachOdd[nextLower[i]];
        }
    }
    
    int count = 0;
    for (int i = 0; i < n; i++) {
        if (canReachOdd[i]) count++;
    }
    
    free(nextHigher);
    free(nextLower);
    free(stack);
    free(indices);
    free(canReachOdd);
    free(canReachEven);
    
    return count;
}

int main() {
    char line[10000];
    fgets(line, sizeof(line), stdin);
    
    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 log n)

Sorting takes O(n log n), monotonic stack operations take O(n), DP takes O(n)

⚡ Linearithmic

Space Complexity

O(n)

Arrays for next jumps, DP states, and stack operations each use O(n) space

⚡ Linearithmic Space

78.0K Views

Medium Frequency

~35 min Avg. Time

1.5K Likes

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Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Dynamic Programming with Monotonic Stack — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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