Minimum Cost to Make Arrays Identical - Practice Coding Problems

Minimum Cost to Make Arrays Identical - Problem

You are given two integer arrays arr and brr of length n, and an integer k.

You can perform the following operations on arr any number of times:

Split and rearrange: Split arr into any number of contiguous subarrays and rearrange these subarrays in any order. This operation has a fixed cost of k.
Adjust element: Choose any element in arr and add or subtract a positive integer x to it. The cost of this operation is x.

Return the minimum total cost to make arr equal to brr.

Input & Output

Example 1 — Basic Case

$ Input: arr = [1,3,2], brr = [2,1,3], k = 1

› Output: 2

💡 Note: Direct adjustment: |1-2| + |3-1| + |2-3| = 4. Sort both arrays: [1,2,3] and [1,2,3], cost = k + 0 = 1. Minimum is 1.

Example 2 — High Rearrangement Cost

$ Input: arr = [2,4,1], brr = [1,2,3], k = 10

› Output: 4

💡 Note: Direct adjustment: |2-1| + |4-2| + |1-3| = 1 + 2 + 2 = 5. Sort cost: k + |1-1| + |2-2| + |4-3| = 10 + 1 = 11. Choose direct adjustment: 5.

Example 3 — Already Optimal

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

› Output: 0

💡 Note: Arrays are already identical, no operations needed. Cost = 0.

Constraints

1 ≤ n ≤ 10⁵
-10⁶ ≤ arr[i], brr[i] ≤ 10⁶
1 ≤ k ≤ 10⁶

Visualization

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Understanding the Visualization

Input

Two arrays arr and brr, and rearrangement cost k

Process

Compare direct adjustment vs sort + adjust strategies

Output

Minimum cost to make arrays identical

Key Takeaway

🎯 Key Insight: Compare direct adjustment cost vs k + optimal pairing cost after sorting

Asked in

G Google 15 a Amazon 12 M Microsoft 8

The key insight is to compare two strategies: direct adjustment vs optimal rearrangement. Best approach is greedy comparison between sorting cost and direct cost. Time: O(n log n), Space: O(n)

Common Approaches

Approach	Time	Space	Notes
✓ Try All Rearrangements	O(2^n × n!)	O(n)	Generate all possible rearrangements and find minimum cost
Compare Two Strategies	O(n log n)	O(n)	Compare cost of sorting once vs adjusting without rearrangement

Try All Rearrangements — Algorithm Steps

Generate all possible subarrays
Try all rearrangements of subarrays
Calculate adjustment cost for each
Return minimum total cost

Visualization

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Step-by-Step Walkthrough

Original Arrays

Compare arr and brr directly

Try Rearrangements

Generate all permutations of arr

Find Minimum

Calculate cost for each and return minimum

Code -

solution.c — C

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

int minCostGlobal = INT_MAX;

int abs_val(int x) {
    return x < 0 ? -x : x;
}

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

void permute(int* arr, int start, int n, int* brr, int k) {
    if (start == n) {
        int cost = k;
        for (int i = 0; i < n; i++) {
            cost += abs_val(arr[i] - brr[i]);
        }
        if (cost < minCostGlobal) {
            minCostGlobal = cost;
        }
        return;
    }
    
    for (int i = start; i < n; i++) {
        swap(&arr[start], &arr[i]);
        permute(arr, start + 1, n, brr, k);
        swap(&arr[start], &arr[i]);
    }
}

int solution(int* arr, int* brr, int n, int k) {
    minCostGlobal = INT_MAX;
    
    // Try without rearrangement
    int cost = 0;
    for (int i = 0; i < n; i++) {
        cost += abs_val(arr[i] - brr[i]);
    }
    if (cost < minCostGlobal) {
        minCostGlobal = cost;
    }
    
    // Try with rearrangement
    int* arrCopy = (int*)malloc(n * sizeof(int));
    for (int i = 0; i < n; i++) {
        arrCopy[i] = arr[i];
    }
    permute(arrCopy, 0, n, brr, k);
    free(arrCopy);
    
    return minCostGlobal;
}

int main() {
    char line[1000];
    
    // Parse arr
    fgets(line, sizeof(line), stdin);
    int arr[100], n = 0;
    char* token = strtok(line, "[,]");
    while (token != NULL) {
        if (token[0] >= '0' && token[0] <= '9' || token[0] == '-') {
            arr[n++] = atoi(token);
        }
        token = strtok(NULL, "[,]");
    }
    
    // Parse brr
    fgets(line, sizeof(line), stdin);
    int brr[100], m = 0;
    token = strtok(line, "[,]");
    while (token != NULL) {
        if (token[0] >= '0' && token[0] <= '9' || token[0] == '-') {
            brr[m++] = atoi(token);
        }
        token = strtok(NULL, "[,]");
    }
    
    int k;
    scanf("%d", &k);
    
    int result = solution(arr, brr, n, k);
    printf("%d\n", result);
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(2^n × n!)

Exponential ways to split array times factorial arrangements

⚠ Quadratic Growth

Space Complexity

O(n)

Space to store current arrangement

⚡ Linearithmic Space

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Input & Output

Constraints

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Related Problems

Common Approaches

Try All Rearrangements — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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