Minimum Element After Replacement With Digit Sum

Minimum Element After Replacement With Digit Sum - Problem

Array Math Easy

You are given an integer array nums. You replace each element in nums with the sum of its digits.

Return the minimum element in nums after all replacements.

Input & Output

Example 1 — Basic Case

$ Input: nums = [10,12,13,14]

› Output: 1

💡 Note: Replace elements: 10→1, 12→3, 13→4, 14→5. Array becomes [1,3,4,5]. Minimum is 1.

Example 2 — Larger Numbers

$ Input: nums = [1,2,3,4]

› Output: 1

💡 Note: Single digit numbers remain unchanged: [1,2,3,4]. Minimum is 1.

Example 3 — Multi-digit Numbers

$ Input: nums = [999,191,22]

› Output: 4

💡 Note: Replace elements: 999→27 (9+9+9), 191→11 (1+9+1), 22→4 (2+2). Array becomes [27,11,4]. Minimum is 4.

Constraints

1 ≤ nums.length ≤ 100
1 ≤ nums[i] ≤ 10⁴

Visualization

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Asked in

M Meta 12 G Google 8

The key insight is to calculate digit sums while tracking the minimum in a single pass. We iterate through each number, compute its digit sum by repeatedly taking modulo 10, and maintain the running minimum. Best approach is the optimized single-pass solution. Time: O(n × d), Space: O(1) where d is average digits per number.

Common Approaches

✓ Brute Force - Two Pass

⏱️ Time: O(n × d) Space: O(1)

First pass through array to calculate digit sum for each element, second pass to find the minimum of all digit sums.

Optimized - Single Pass

⏱️ Time: O(n × d) Space: O(1)

Instead of transforming the entire array first, we calculate digit sum for each element and immediately compare with current minimum, avoiding the need for a second pass.

Brute Force - Two Pass — Algorithm Steps

Step 1: For each number, calculate sum of its digits
Step 2: Replace original number with digit sum
Step 3: Find minimum in the modified array

Visualization

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Step-by-Step Walkthrough

Original Array

Input array with original values

Calculate Digit Sums

Replace each element with sum of digits

Find Minimum

Scan transformed array for minimum value

Code -

solution.c — C

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

int digitSum(int num) {
    num = abs(num);
    int total = 0;
    while (num > 0) {
        total += num % 10;
        num /= 10;
    }
    return total;
}

int solution(int* nums, int numsSize) {
    // Replace each element with digit sum
    for (int i = 0; i < numsSize; i++) {
        nums[i] = digitSum(nums[i]);
    }
    
    // Find minimum
    int min = nums[0];
    for (int i = 1; i < numsSize; i++) {
        if (nums[i] < min) {
            min = nums[i];
        }
    }
    return min;
}

int main() {
    char line[10000];
    fgets(line, sizeof(line), stdin);
    
    // Simple parsing for [num,num,num] format
    int nums[1000];
    int count = 0;
    char* token = strtok(line, "[,]");
    while (token != NULL) {
        nums[count++] = atoi(token);
        token = strtok(NULL, "[,]");
    }
    
    int result = solution(nums, count);
    printf("%d\n", result);
    
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n × d)

n elements, each with up to d digits to sum

✓ Linear Growth

Space Complexity

O(1)

Only using variables for calculation, modifying array in-place

✓ Linear Space

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Input & Output

Constraints

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Related Problems

Common Approaches

Brute Force - Two Pass — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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