LeetCode #2578 — EASY

Split With Minimum Sum

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

Solve on LeetCode

The Problem

Problem Statement

Given a positive integer num, split it into two non-negative integers num1 and num2 such that:

The concatenation of num1 and num2 is a permutation of num.
- In other words, the sum of the number of occurrences of each digit in num1 and num2 is equal to the number of occurrences of that digit in num.
num1 and num2 can contain leading zeros.

Return the minimum possible sum of num1 and num2.

Notes:

It is guaranteed that num does not contain any leading zeros.
The order of occurrence of the digits in num1 and num2 may differ from the order of occurrence of num.

Example 1:

Input: num = 4325
Output: 59
Explanation: We can split 4325 so that num1 is 24 and num2 is 35, giving a sum of 59. We can prove that 59 is indeed the minimal possible sum.

Example 2:

Input: num = 687
Output: 75
Explanation: We can split 687 so that num1 is 68 and num2 is 7, which would give an optimal sum of 75.

Constraints:

10 <= num <= 10⁹

Step 01

Brute Force Baseline

Problem summary: Given a positive integer num, split it into two non-negative integers num1 and num2 such that: The concatenation of num1 and num2 is a permutation of num. In other words, the sum of the number of occurrences of each digit in num1 and num2 is equal to the number of occurrences of that digit in num. num1 and num2 can contain leading zeros. Return the minimum possible sum of num1 and num2. Notes: It is guaranteed that num does not contain any leading zeros. The order of occurrence of the digits in num1 and num2 may differ from the order of occurrence of num.

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

Sort the digits of num in non decreasing order.
Assign digits to num1 and num2 alternatively.

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 #2578: Split With Minimum Sum
class Solution {
    public int splitNum(int num) {
        int[] cnt = new int[10];
        int n = 0;
        for (; num > 0; num /= 10) {
            ++cnt[num % 10];
            ++n;
        }
        int[] ans = new int[2];
        for (int i = 0, j = 0; i < n; ++i) {
            while (cnt[j] == 0) {
                ++j;
            }
            --cnt[j];
            ans[i & 1] = ans[i & 1] * 10 + j;
        }
        return ans[0] + ans[1];
    }
}

// Accepted solution for LeetCode #2578: Split With Minimum Sum
func splitNum(num int) int {
	cnt := [10]int{}
	n := 0
	for ; num > 0; num /= 10 {
		cnt[num%10]++
		n++
	}
	ans := [2]int{}
	for i, j := 0, 0; i < n; i++ {
		for cnt[j] == 0 {
			j++
		}
		cnt[j]--
		ans[i&1] = ans[i&1]*10 + j
	}
	return ans[0] + ans[1]
}

# Accepted solution for LeetCode #2578: Split With Minimum Sum
class Solution:
    def splitNum(self, num: int) -> int:
        cnt = Counter()
        n = 0
        while num:
            cnt[num % 10] += 1
            num //= 10
            n += 1
        ans = [0] * 2
        j = 0
        for i in range(n):
            while cnt[j] == 0:
                j += 1
            cnt[j] -= 1
            ans[i & 1] = ans[i & 1] * 10 + j
        return sum(ans)

// Accepted solution for LeetCode #2578: Split With Minimum Sum
impl Solution {
    pub fn split_num(mut num: i32) -> i32 {
        let mut cnt = vec![0; 10];
        let mut n = 0;

        while num != 0 {
            cnt[(num as usize) % 10] += 1;
            num /= 10;
            n += 1;
        }

        let mut ans = vec![0; 2];
        let mut j = 0;
        for i in 0..n {
            while cnt[j] == 0 {
                j += 1;
            }
            cnt[j] -= 1;

            ans[i & 1] = ans[i & 1] * 10 + (j as i32);
        }

        ans[0] + ans[1]
    }
}

// Accepted solution for LeetCode #2578: Split With Minimum Sum
function splitNum(num: number): number {
    const cnt: number[] = Array(10).fill(0);
    let n = 0;
    for (; num > 0; num = Math.floor(num / 10)) {
        ++cnt[num % 10];
        ++n;
    }
    const ans: number[] = Array(2).fill(0);
    for (let i = 0, j = 0; i < n; ++i) {
        while (cnt[j] === 0) {
            ++j;
        }
        --cnt[j];
        ans[i & 1] = ans[i & 1] * 10 + j;
    }
    return ans[0] + ans[1];
}

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)

Space

O(C)

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.