LeetCode #2165 — MEDIUM

Smallest Value of the Rearranged Number

Move from brute-force thinking to an efficient approach using math strategy.

The Problem

Problem Statement

You are given an integer num. Rearrange the digits of num such that its value is minimized and it does not contain any leading zeros.

Return the rearranged number with minimal value.

Note that the sign of the number does not change after rearranging the digits.

Example 1:

Input: num = 310
Output: 103
Explanation: The possible arrangements for the digits of 310 are 013, 031, 103, 130, 301, 310. 
The arrangement with the smallest value that does not contain any leading zeros is 103.

Example 2:

Input: num = -7605
Output: -7650
Explanation: Some possible arrangements for the digits of -7605 are -7650, -6705, -5076, -0567.
The arrangement with the smallest value that does not contain any leading zeros is -7650.

Constraints:

-10¹⁵ <= num <= 10¹⁵

Step 01

Brute Force Baseline

Problem summary: You are given an integer num. Rearrange the digits of num such that its value is minimized and it does not contain any leading zeros. Return the rearranged number with minimal value. Note that the sign of the number does not change after rearranging the digits.

Baseline thinking

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

Pattern signal: Math

Example 1

Example 2

-7605

Core Insight

What unlocks the optimal approach

For positive numbers, the leading digit should be the smallest nonzero digit. Then the remaining digits follow in ascending order.
For negative numbers, the digits should be arranged in descending order.

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 #2165: Smallest Value of the Rearranged Number
class Solution {
    public long smallestNumber(long num) {
        boolean neg = num < 0;
        num = Math.abs(num);
        int[] cnt = new int[10];
        while (num > 0) {
            ++cnt[(int) (num % 10)];
            num /= 10;
        }
        long ans = 0;
        if (neg) {
            for (int i = 9; i >= 0; --i) {
                while (cnt[i] > 0) {
                    ans = ans * 10 + i;
                    --cnt[i];
                }
            }
            return -ans;
        }
        if (cnt[0] > 0) {
            for (int i = 1; i < 10; ++i) {
                if (cnt[i] > 0) {
                    --cnt[i];
                    ans = i;
                    break;
                }
            }
        }
        for (int i = 0; i < 10; ++i) {
            while (cnt[i] > 0) {
                ans = ans * 10 + i;
                --cnt[i];
            }
        }
        return ans;
    }
}

// Accepted solution for LeetCode #2165: Smallest Value of the Rearranged Number
func smallestNumber(num int64) (ans int64) {
	neg := num < 0
	num = max(num, -num)
	cnt := make([]int, 10)

	for num > 0 {
		cnt[num%10]++
		num /= 10
	}

	if neg {
		for i := 9; i >= 0; i-- {
			for cnt[i] > 0 {
				ans = ans*10 + int64(i)
				cnt[i]--
			}
		}
		return -ans
	}

	if cnt[0] > 0 {
		for i := 1; i < 10; i++ {
			if cnt[i] > 0 {
				cnt[i]--
				ans = int64(i)
				break
			}
		}
	}

	for i := 0; i < 10; i++ {
		for cnt[i] > 0 {
			ans = ans*10 + int64(i)
			cnt[i]--
		}
	}

	return ans
}

# Accepted solution for LeetCode #2165: Smallest Value of the Rearranged Number
class Solution:
    def smallestNumber(self, num: int) -> int:
        neg = num < 0
        num = abs(num)
        cnt = [0] * 10
        while num:
            cnt[num % 10] += 1
            num //= 10
        ans = 0
        if neg:
            for i in reversed(range(10)):
                for _ in range(cnt[i]):
                    ans *= 10
                    ans += i
            return -ans
        if cnt[0]:
            for i in range(1, 10):
                if cnt[i]:
                    ans = i
                    cnt[i] -= 1
                    break
        for i in range(10):
            for _ in range(cnt[i]):
                ans *= 10
                ans += i
        return ans

// Accepted solution for LeetCode #2165: Smallest Value of the Rearranged Number
impl Solution {
    pub fn smallest_number(num: i64) -> i64 {
        let mut neg = num < 0;
        let mut num = num.abs();
        let mut cnt = vec![0; 10];

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

        let mut ans = 0;
        if neg {
            for i in (0..10).rev() {
                while cnt[i] > 0 {
                    ans = ans * 10 + i as i64;
                    cnt[i] -= 1;
                }
            }
            return -ans;
        }

        if cnt[0] > 0 {
            for i in 1..10 {
                if cnt[i] > 0 {
                    cnt[i] -= 1;
                    ans = i as i64;
                    break;
                }
            }
        }

        for i in 0..10 {
            while cnt[i] > 0 {
                ans = ans * 10 + i as i64;
                cnt[i] -= 1;
            }
        }

        ans
    }
}

// Accepted solution for LeetCode #2165: Smallest Value of the Rearranged Number
function smallestNumber(num: number): number {
    const neg = num < 0;
    num = Math.abs(num);
    const cnt = Array(10).fill(0);

    while (num > 0) {
        cnt[num % 10]++;
        num = Math.floor(num / 10);
    }

    let ans = 0;
    if (neg) {
        for (let i = 9; i >= 0; i--) {
            while (cnt[i] > 0) {
                ans = ans * 10 + i;
                cnt[i]--;
            }
        }
        return -ans;
    }

    if (cnt[0] > 0) {
        for (let i = 1; i < 10; i++) {
            if (cnt[i] > 0) {
                cnt[i]--;
                ans = i;
                break;
            }
        }
    }

    for (let i = 0; i < 10; i++) {
        while (cnt[i] > 0) {
            ans = ans * 10 + i;
            cnt[i]--;
        }
    }

    return ans;
}

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(log n)

Space

O(1)

Approach Breakdown

ITERATIVE

O(n) time

O(1) space

Simulate the process step by step — multiply n times, check each number up to n, or iterate through all possibilities. Each step is O(1), but doing it n times gives O(n). No extra space needed since we just track running state.

MATH INSIGHT

O(log n) time

O(1) space

Math problems often have a closed-form or O(log n) solution hidden behind an O(n) simulation. Modular arithmetic, fast exponentiation (repeated squaring), GCD (Euclidean algorithm), and number theory properties can dramatically reduce complexity.

Shortcut: Look for mathematical properties that eliminate iteration. Repeated squaring → O(log n). Modular arithmetic avoids overflow.

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.