Maximum Prime Difference

Maximum Prime Difference - Problem

You are given an integer array nums. Return an integer that is the maximum distance between the indices of two (not necessarily different) prime numbers in nums.

The distance between two indices i and j is |i - j|.

A prime number is a natural number greater than 1 that has no positive divisors other than 1 and itself.

Input & Output

Example 1 — Multiple Primes

$ Input: nums = [4,2,9,5,3]

› Output: 3

💡 Note: Prime numbers are 2 (index 1), 5 (index 3), and 3 (index 4). Maximum distance is between indices 1 and 4: |4-1| = 3.

Example 2 — Same Prime Twice

$ Input: nums = [4,8,2,8]

› Output: 0

💡 Note: Only one prime number 2 at index 2. Distance between the same index is 0.

Example 3 — Edge Positions

$ Input: nums = [2,4,6,8,11]

› Output: 4

💡 Note: Primes are 2 (index 0) and 11 (index 4). Maximum distance is |4-0| = 4.

Constraints

1 ≤ nums.length ≤ 3 × 10⁵
1 ≤ nums[i] ≤ 10⁶

Visualization

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Asked in

G Google 15 a Amazon 12 M Microsoft 8

The key insight is that the maximum distance between any two prime positions is always between the first and last prime in the array. Best approach uses a single pass to track first and last prime positions. Time: O(n × √max), Space: O(1)

Common Approaches

Approach	Time	Space	Notes
✓ Brute Force - Check All Prime Pairs	O(n × √max + p²)	O(p)	Find all prime numbers and check distance between every pair
Single Pass - First and Last Prime	O(n × √max)	O(1)	Track only the first and last prime positions in one scan

Brute Force - Check All Prime Pairs — Algorithm Steps

Find all positions with prime numbers
Check distance between every pair of prime positions
Return the maximum distance found

Visualization

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Step-by-Step Walkthrough

Find Primes

Identify all prime numbers and their positions

Check Pairs

Calculate distance between every pair of primes

Max Distance

Return the largest distance found

Code -

solution.c — C

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

int isPrime(int n) {
    if (n < 2) return 0;
    if (n == 2) return 1;
    if (n % 2 == 0) return 0;
    for (int i = 3; i * i <= n; i += 2) {
        if (n % i == 0) return 0;
    }
    return 1;
}

int solution(int* nums, int numsSize) {
    int primeIndices[1000];
    int primeCount = 0;
    
    // Find all prime positions
    for (int i = 0; i < numsSize; i++) {
        if (isPrime(nums[i])) {
            primeIndices[primeCount++] = i;
        }
    }
    
    int maxDistance = 0;
    // Check all pairs of prime positions
    for (int i = 0; i < primeCount; i++) {
        for (int j = i + 1; j < primeCount; j++) {
            int distance = primeIndices[j] - primeIndices[i];
            if (distance > maxDistance) {
                maxDistance = distance;
            }
        }
    }
    
    return maxDistance;
}

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

Time & Space Complexity

Time Complexity

⏱️

O(n × √max + p²)

O(n × √max) for prime checking each element, O(p²) for comparing all prime pairs where p is number of primes

✓ Linear Growth

Space Complexity

O(p)

Store positions of all prime numbers in array

✓ Linear Space

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Input & Output

Constraints

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Related Problems

Common Approaches

Brute Force - Check All Prime Pairs — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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