Smallest Range II

Smallest Range II - Problem

You are given an integer array nums and an integer k. For each index i where 0 <= i < nums.length, change nums[i] to be either nums[i] + k or nums[i] - k.

The score of nums is the difference between the maximum and minimum elements in nums.

Return the minimum score of nums after changing the values at each index.

Input & Output

Example 1 — Basic Case

$ Input: nums = [1,3,6], k = 3

› Output: 3

💡 Note: After changes: [1+3, 3+3, 6-3] = [4,6,3]. Range = max(4,6,3) - min(4,6,3) = 6-3 = 3

Example 2 — All Same Operation

$ Input: nums = [0,10], k = 2

› Output: 6

💡 Note: Best choice: [0+2, 10-2] = [2,8]. Range = 8-2 = 6. (Better than [2,12] range=10 or [-2,8] range=10)

Example 3 — Single Element

$ Input: nums = [1], k = 0

› Output: 0

💡 Note: Only one element, so range is always 0 regardless of operations

Constraints

1 ≤ nums.length ≤ 10⁴
-10⁴ ≤ nums[i] ≤ 10⁴
0 ≤ k ≤ 10⁴

Visualization

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Asked in

G Google 25 f Facebook 18 a Amazon 12

The key insight is that after sorting, the optimal solution has a prefix-suffix structure where elements before some split point add k and elements after subtract k. The greedy approach sorts the array and tries all possible split points in O(n log n) time with O(1) space.

Common Approaches

✓ Brute Force - Try All Combinations

⏱️ Time: O(2ⁿ × n) Space: O(n)

For each element, we can either add k or subtract k. Generate all possible combinations and find the one with minimum range.

Greedy Optimization

⏱️ Time: O(n log n) Space: O(1)

After sorting, we can prove that there exists an optimal solution where some prefix of elements gets +k and the suffix gets -k. We try all possible split points.

Sort and Split Strategy

⏱️ Time: O(n log n) Space: O(1)

Sort the array first, then try each possible split point where elements before add k and elements after subtract k. The key insight is that the optimal solution will have this structure.

Brute Force - Try All Combinations — Algorithm Steps

Generate all 2^n combinations of +k/-k operations
Calculate max-min for each combination
Return the minimum range found

Visualization

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Step-by-Step Walkthrough

Input

Array [1,3,6] with k=3

Generate

Try all 2³=8 combinations of +3/-3

Result

Find combination with minimum range

Code -

solution.c — C

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

int solution(int* nums, int numsSize, int k) {
    int minRange = INT_MAX;
    
    // Try all 2^n combinations
    for (int mask = 0; mask < (1 << numsSize); mask++) {
        int* arr = (int*)malloc(numsSize * sizeof(int));
        for (int i = 0; i < numsSize; i++) {
            if (mask & (1 << i)) {
                arr[i] = nums[i] + k;
            } else {
                arr[i] = nums[i] - k;
            }
        }
        
        int maxVal = arr[0], minVal = arr[0];
        for (int i = 1; i < numsSize; i++) {
            if (arr[i] > maxVal) maxVal = arr[i];
            if (arr[i] < minVal) minVal = arr[i];
        }
        
        int currentRange = maxVal - minVal;
        if (currentRange < minRange) {
            minRange = currentRange;
        }
        
        free(arr);
    }
    
    return minRange;
}

int main() {
    char line[1000];
    fgets(line, sizeof(line), stdin);
    
    // Parse array
    int nums[100];
    int numsSize = 0;
    char* token = strtok(line, "[],\n");
    while (token != NULL) {
        if (strlen(token) > 0) {
            nums[numsSize++] = atoi(token);
        }
        token = strtok(NULL, "[],\n");
    }
    
    int k;
    scanf("%d", &k);
    
    printf("%d\n", solution(nums, numsSize, k));
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(2ⁿ × n)

2^n combinations, each takes O(n) to compute min/max

⚡ Linearithmic

Space Complexity

O(n)

Array to store current combination

✓ Linear Space

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Smart Actions

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Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Brute Force - Try All Combinations — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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