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Advantage Shuffle - Problem

Advantage Shuffle - The Strategic Card Game Problem

Imagine you're playing a card game where you need to maximize your wins! You have two decks of cards with numbers: nums1 (your deck) and nums2 (opponent's deck). The goal is to rearrange your deck to beat as many of your opponent's cards as possible.

🎯 The Rules:
• You win a round when your card value > opponent's card value at the same position
• Your advantage is the total number of rounds you win
• You can shuffle your deck in any order, but the opponent's deck stays fixed

Input: Two integer arrays of equal length
Output: Any permutation of nums1 that maximizes the number of positions where nums1[i] > nums2[i]

Example: If your cards are [2,7,11,15] and opponent has [1,10,4,11], you could arrange yours as [2,11,7,15] to win 3 out of 4 rounds!

Input & Output

example_1.py — Basic Case

$ Input: nums1 = [2,7,11,15], nums2 = [1,10,4,11]

› Output: [2,11,7,15]

💡 Note: We can beat all opponent cards! 2>1, 11>10, 7>4, 15>11. This gives us maximum advantage of 4.

example_2.py — Strategic Sacrifice

$ Input: nums1 = [12,24,8,32], nums2 = [13,25,32,11]

› Output: [24,32,8,12]

💡 Note: Strategic assignment: 24>13, 32>25, sacrifice 8 to 32, 12>11. Total advantage: 3 out of 4.

example_3.py — Edge Case

$ Input: nums1 = [1,2], nums2 = [3,4]

› Output: [1,2] or [2,1]

💡 Note: No card can beat any opponent card, so any arrangement gives 0 advantage. We can return any permutation.

Constraints

1 ≤ nums1.length ≤ 10⁵
nums2.length == nums1.length
0 ≤ nums1[i], nums2[i] ≤ 10⁹
Both arrays contain the same number of elements

Visualization

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Understanding the Visualization

Intelligence Gathering

Sort your cards and opponent's cards to understand the battlefield

Strategic Assignment

For tough opponent cards, use your best available card that can win

Tactical Sacrifice

When you can't win, sacrifice your weakest card to save stronger ones

Key Takeaway

🎯 Key Insight: The greedy strategy of using the smallest card that can win (or sacrificing the smallest when losing) maximizes our advantage by preserving stronger cards for tougher future battles!

Asked in

G Google 35 a Amazon 28 f Meta 22 ⊞ Microsoft 18

The optimal solution uses a greedy two-pointers approach with O(n log n) time complexity. Key insight: sort both arrays and for each opponent card (processed from largest to smallest), either use our smallest winning card or sacrifice our smallest card when we can't win. This maximizes future winning opportunities and guarantees the maximum possible advantage.

Common Approaches

Approach	Time	Space	Notes
✓ Brute Force (Try All Permutations)	O(n! × n)	O(n)	Generate all possible permutations of nums1 and find the one with maximum advantage
Sorting + Binary Search	O(n log n)	O(n)	Sort nums1 and for each nums2 element, binary search for the smallest winning card
Two Pointers with Greedy Strategy (Optimal)	O(n log n)	O(n)	Sort both arrays and use greedy matching with two pointers to maximize advantage

Brute Force (Try All Permutations) — Algorithm Steps

Generate all possible permutations of nums1
For each permutation, compare with nums2 position by position
Count how many positions where nums1[i] > nums2[i]
Track the permutation with maximum advantage
Return the best permutation found

Visualization

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Step-by-Step Walkthrough

Generate Permutations

Create all possible orderings of nums1

Calculate Advantage

For each permutation, count wins against nums2

Track Best

Keep the permutation with highest advantage

Code -

solution.c — C

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

int calculateAdvantage(int* perm, int* nums2, int n) {
    int count = 0;
    for (int i = 0; i < n; i++) {
        if (perm[i] > nums2[i]) count++;
    }
    return count;
}

void swap(int* a, int* b) {
    int temp = *a;
    *a = *b;
    *b = temp;
}

void permute(int* arr, int start, int n, int* nums2, int** bestResult, int* bestAdvantage) {
    if (start == n) {
        int advantage = calculateAdvantage(arr, nums2, n);
        if (advantage > *bestAdvantage) {
            *bestAdvantage = advantage;
            memcpy(*bestResult, arr, n * sizeof(int));
        }
        return;
    }
    
    for (int i = start; i < n; i++) {
        swap(&arr[start], &arr[i]);
        permute(arr, start + 1, n, nums2, bestResult, bestAdvantage);
        swap(&arr[start], &arr[i]);
    }
}

int* advantageCount(int* nums1, int nums1Size, int* nums2, int nums2Size, int* returnSize) {
    *returnSize = nums1Size;
    int* result = (int*)malloc(nums1Size * sizeof(int));
    memcpy(result, nums1, nums1Size * sizeof(int));
    
    int bestAdvantage = calculateAdvantage(nums1, nums2, nums1Size);
    int* bestResult = result;
    
    permute(nums1, 0, nums1Size, nums2, &bestResult, &bestAdvantage);
    return result;
}

int main() {
    // Simple input parsing
    int nums1[] = {2, 7, 11, 15};
    int nums2[] = {1, 10, 4, 11};
    int returnSize;
    
    int* result = advantageCount(nums1, 4, nums2, 4, &returnSize);
    
    printf("[");
    for (int i = 0; i < returnSize; i++) {
        if (i > 0) printf(",");
        printf("%d", result[i]);
    }
    printf("]\n");
    
    free(result);
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n! × n)

n! permutations, each taking O(n) time to evaluate advantage

⚠ Quadratic Growth

Space Complexity

O(n)

Space for storing current permutation and tracking best result

⚡ Linearithmic Space

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Medium Frequency

~18 min Avg. Time

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Smart Actions

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Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Brute Force (Try All Permutations) — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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