Off-by-one on range boundaries
Wrong move: Loop endpoints miss first/last candidate.
Usually fails on: Fails on minimal arrays and exact-boundary answers.
Fix: Re-derive loops from inclusive/exclusive ranges before coding.
Build confidence with an intuition-first walkthrough focused on array fundamentals.
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.
Example 1:
Input: nums = [10,12,13,14]
Output: 1
Explanation:
nums becomes [1, 3, 4, 5] after all replacements, with minimum element 1.
Example 2:
Input: nums = [1,2,3,4]
Output: 1
Explanation:
nums becomes [1, 2, 3, 4] after all replacements, with minimum element 1.
Example 3:
Input: nums = [999,19,199]
Output: 10
Explanation:
nums becomes [27, 10, 19] after all replacements, with minimum element 10.
Constraints:
1 <= nums.length <= 1001 <= nums[i] <= 104Problem summary: 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.
Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.
Pattern signal: Array · Math
[10,12,13,14]
[1,2,3,4]
[999,19,199]
sum-of-digits-of-string-after-convert)Source-backed implementations are provided below for direct study and interview prep.
// Accepted solution for LeetCode #3300: Minimum Element After Replacement With Digit Sum
class Solution {
public int minElement(int[] nums) {
int ans = 100;
for (int x : nums) {
int y = 0;
for (; x > 0; x /= 10) {
y += x % 10;
}
ans = Math.min(ans, y);
}
return ans;
}
}
// Accepted solution for LeetCode #3300: Minimum Element After Replacement With Digit Sum
func minElement(nums []int) int {
ans := 100
for _, x := range nums {
y := 0
for ; x > 0; x /= 10 {
y += x % 10
}
ans = min(ans, y)
}
return ans
}
# Accepted solution for LeetCode #3300: Minimum Element After Replacement With Digit Sum
class Solution:
def minElement(self, nums: List[int]) -> int:
return min(sum(int(b) for b in str(x)) for x in nums)
// Accepted solution for LeetCode #3300: Minimum Element After Replacement With Digit Sum
fn min_element(nums: Vec<i32>) -> i32 {
nums.into_iter()
.map(|n| {
let mut n = n;
let mut sum = 0;
while n > 0 {
sum += n % 10;
n /= 10;
}
sum
})
.min()
.unwrap()
}
fn main() {
let nums = vec![10, 12, 13, 14];
let ret = min_element(nums);
println!("ret={ret}");
}
#[test]
fn test() {
{
let nums = vec![10, 12, 13, 14];
let ret = min_element(nums);
assert_eq!(ret, 1);
}
{
let nums = vec![999, 19, 199];
let ret = min_element(nums);
assert_eq!(ret, 10);
}
{
let nums = vec![1, 2, 3, 4];
let ret = min_element(nums);
assert_eq!(ret, 1);
}
}
// Accepted solution for LeetCode #3300: Minimum Element After Replacement With Digit Sum
function minElement(nums: number[]): number {
let ans: number = 100;
for (let x of nums) {
let y = 0;
for (; x; x = Math.floor(x / 10)) {
y += x % 10;
}
ans = Math.min(ans, y);
}
return ans;
}
Use this to step through a reusable interview workflow for this problem.
Two nested loops check every pair or subarray. The outer loop fixes a starting point, the inner loop extends or searches. For n elements this gives up to n²/2 operations. No extra space, but the quadratic time is prohibitive for large inputs.
Most array problems have an O(n²) brute force (nested loops) and an O(n) optimal (single pass with clever state tracking). The key is identifying what information to maintain as you scan: a running max, a prefix sum, a hash map of seen values, or two pointers.
Review these before coding to avoid predictable interview regressions.
Wrong move: Loop endpoints miss first/last candidate.
Usually fails on: Fails on minimal arrays and exact-boundary answers.
Fix: Re-derive loops from inclusive/exclusive ranges before coding.
Wrong move: Temporary multiplications exceed integer bounds.
Usually fails on: Large inputs wrap around unexpectedly.
Fix: Use wider types, modular arithmetic, or rearranged operations.