LeetCode #2160 — EASY

Minimum Sum of Four Digit Number After Splitting Digits

Build confidence with an intuition-first walkthrough focused on math fundamentals.

The Problem

Problem Statement

You are given a positive integer num consisting of exactly four digits. Split num into two new integers new1 and new2 by using the digits found in num. Leading zeros are allowed in new1 and new2, and all the digits found in num must be used.

For example, given num = 2932, you have the following digits: two 2's, one 9 and one 3. Some of the possible pairs [new1, new2] are [22, 93], [23, 92], [223, 9] and [2, 329].

Return the minimum possible sum of new1 and new2.

Example 1:

Input: num = 2932
Output: 52
Explanation: Some possible pairs [new1, new2] are [29, 23], [223, 9], etc.
The minimum sum can be obtained by the pair [29, 23]: 29 + 23 = 52.

Example 2:

Input: num = 4009
Output: 13
Explanation: Some possible pairs [new1, new2] are [0, 49], [490, 0], etc. 
The minimum sum can be obtained by the pair [4, 9]: 4 + 9 = 13.

Constraints:

1000 <= num <= 9999

Step 01

Brute Force Baseline

Problem summary: You are given a positive integer num consisting of exactly four digits. Split num into two new integers new1 and new2 by using the digits found in num. Leading zeros are allowed in new1 and new2, and all the digits found in num must be used. For example, given num = 2932, you have the following digits: two 2's, one 9 and one 3. Some of the possible pairs [new1, new2] are [22, 93], [23, 92], [223, 9] and [2, 329]. Return the minimum possible sum of new1 and new2.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Math · Greedy

Example 1

Example 2

Core Insight

What unlocks the optimal approach

Notice that the most optimal way to obtain the minimum possible sum using 4 digits is by summing up two 2-digit numbers.
We can use the two smallest digits out of the four as the digits found in the tens place respectively.
Similarly, we use the final 2 larger digits as the digits found in the ones place.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #2160: Minimum Sum of Four Digit Number After Splitting Digits
class Solution {
    public int minimumSum(int num) {
        int[] nums = new int[4];
        for (int i = 0; num != 0; ++i) {
            nums[i] = num % 10;
            num /= 10;
        }
        Arrays.sort(nums);
        return 10 * (nums[0] + nums[1]) + nums[2] + nums[3];
    }
}

// Accepted solution for LeetCode #2160: Minimum Sum of Four Digit Number After Splitting Digits
func minimumSum(num int) int {
	var nums []int
	for num > 0 {
		nums = append(nums, num%10)
		num /= 10
	}
	sort.Ints(nums)
	return 10*(nums[0]+nums[1]) + nums[2] + nums[3]
}

# Accepted solution for LeetCode #2160: Minimum Sum of Four Digit Number After Splitting Digits
class Solution:
    def minimumSum(self, num: int) -> int:
        nums = []
        while num:
            nums.append(num % 10)
            num //= 10
        nums.sort()
        return 10 * (nums[0] + nums[1]) + nums[2] + nums[3]

// Accepted solution for LeetCode #2160: Minimum Sum of Four Digit Number After Splitting Digits
impl Solution {
    pub fn minimum_sum(mut num: i32) -> i32 {
        let mut nums = [0; 4];
        for i in 0..4 {
            nums[i] = num % 10;
            num /= 10;
        }
        nums.sort();
        10 * (nums[0] + nums[1]) + nums[2] + nums[3]
    }
}

// Accepted solution for LeetCode #2160: Minimum Sum of Four Digit Number After Splitting Digits
function minimumSum(num: number): number {
    const nums = new Array(4).fill(0);
    for (let i = 0; i < 4; i++) {
        nums[i] = num % 10;
        num = Math.floor(num / 10);
    }
    nums.sort((a, b) => a - b);
    return 10 * (nums[0] + nums[1]) + nums[2] + nums[3];
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n log n)

Space

O(1)

Approach Breakdown

EXHAUSTIVE

O(2ⁿ) time

O(n) space

Try every possible combination of choices. With n items each having two states (include/exclude), the search space is 2ⁿ. Evaluating each combination takes O(n), giving O(n × 2ⁿ). The recursion stack or subset storage uses O(n) space.

GREEDY

O(n log n) time

O(1) space

Greedy algorithms typically sort the input (O(n log n)) then make a single pass (O(n)). The sort dominates. If the input is already sorted or the greedy choice can be computed without sorting, time drops to O(n). Proving greedy correctness (exchange argument) is harder than the implementation.

Shortcut: Sort + single pass → O(n log n). If no sort needed → O(n). The hard part is proving it works.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Overflow in intermediate arithmetic

Wrong move: Temporary multiplications exceed integer bounds.

Usually fails on: Large inputs wrap around unexpectedly.

Fix: Use wider types, modular arithmetic, or rearranged operations.

Using greedy without proof

Wrong move: Locally optimal choices may fail globally.

Usually fails on: Counterexamples appear on crafted input orderings.

Fix: Verify with exchange argument or monotonic objective before committing.