Jump Game IV - Practice Coding Problems

Jump Game IV - Problem

Given an array of integers arr, you are initially positioned at the first index of the array.

In one step you can jump from index i to index:

i + 1 where: i + 1 < arr.length.
i - 1 where: i - 1 >= 0.
j where: arr[i] == arr[j] and i != j.

Return the minimum number of steps to reach the last index of the array.

Notice that you can not jump outside of the array at any time.

Input & Output

Example 1 — Basic Teleportation

$ Input: arr = [100,-23,-23,404,100,23,23,23,3,404]

› Output: 3

💡 Note: Start at index 0 (100) → Jump to index 4 (same value 100) → Jump to index 5 (23) → Jump to index 9 (404, last index). Total: 3 steps.

Example 2 — Adjacent Jumps

$ Input: arr = [7]

› Output: 0

💡 Note: Already at the last index, so 0 steps needed.

Example 3 — Mixed Strategy

$ Input: arr = [7,6,9,6,9,6,9,7]

› Output: 1

💡 Note: Start at index 0 (7) → Jump directly to index 7 (same value 7, which is the last index). Total: 1 step.

Constraints

1 ≤ arr.length ≤ 5 × 10⁴
-10⁸ ≤ arr[i] ≤ 10⁸

Visualization

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Understanding the Visualization

Input Array

Array with indices 0 to n-1, goal is to reach last index

Jump Types

Left (-1), Right (+1), or Teleport (same value)

Minimum Steps

Find shortest path using BFS

Key Takeaway

🎯 Key Insight: BFS with value-to-indices mapping enables O(1) teleportation lookups for optimal O(n) solution

Asked in

f Facebook 15 a Amazon 12 G Google 8 M Microsoft 6

The key insight is to use BFS for guaranteed minimum steps while preprocessing values into a map for O(1) teleportation lookups. The optimal approach uses BFS with value mapping achieving O(n) time and O(n) space complexity.

Common Approaches

Approach	Time	Space	Notes
✓ DFS with Memoization	O(n²)	O(n)	Recursively try all possible jumps with memoization
Breadth-First Search	O(n²)	O(n)	BFS to find minimum steps level by level
BFS with Value Mapping	O(n)	O(n)	BFS with preprocessing to group indices by value

DFS with Memoization — Algorithm Steps

Start from index 0
For each index, try all three jump types
Recursively solve for each reachable position
Return minimum steps among all possibilities

Visualization

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Step-by-Step Walkthrough

Start at Index 0

Begin DFS from first position

Try All Jumps

Explore left, right, and same-value jumps

Memoize Results

Cache minimum steps for each index

Code -

solution.c — C

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

#define MAX_SIZE 1000

int memo[MAX_SIZE];
int arr[MAX_SIZE];
int n;
int memo_set[MAX_SIZE];

int dfs(int i) {
    if (i == n - 1) return 0;
    if (memo_set[i]) return memo[i];
    
    int result = INT_MAX;
    
    // Jump right
    if (i + 1 < n) {
        int steps = 1 + dfs(i + 1);
        if (steps < result) result = steps;
    }
    
    // Jump left
    if (i - 1 >= 0) {
        int steps = 1 + dfs(i - 1);
        if (steps < result) result = steps;
    }
    
    // Jump to same value
    for (int j = 0; j < n; j++) {
        if (j != i && arr[j] == arr[i]) {
            int steps = 1 + dfs(j);
            if (steps < result) result = steps;
        }
    }
    
    memo[i] = result;
    memo_set[i] = 1;
    return result;
}

int solution(int* array, int size) {
    n = size;
    if (n <= 1) return 0;
    
    for (int i = 0; i < n; i++) {
        arr[i] = array[i];
        memo_set[i] = 0;
    }
    
    return dfs(0);
}

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

Time & Space Complexity

Time Complexity

⏱️

O(n²)

In worst case, we visit each index and for each index check all other indices with same value

⚠ Quadratic Growth

Space Complexity

O(n)

Memoization array and recursion call stack depth up to n

⚡ Linearithmic Space

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Smart Actions

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Input & Output

Constraints

Visualization

Related Problems

Common Approaches

DFS with Memoization — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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