LeetCode #2442 — MEDIUM

Count Number of Distinct Integers After Reverse Operations

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

The Problem

Problem Statement

You are given an array nums consisting of positive integers.

You have to take each integer in the array, reverse its digits, and add it to the end of the array. You should apply this operation to the original integers in nums.

Return the number of distinct integers in the final array.

Example 1:

Input: nums = [1,13,10,12,31]
Output: 6
Explanation: After including the reverse of each number, the resulting array is [1,13,10,12,31,1,31,1,21,13].
The reversed integers that were added to the end of the array are underlined. Note that for the integer 10, after reversing it, it becomes 01 which is just 1.
The number of distinct integers in this array is 6 (The numbers 1, 10, 12, 13, 21, and 31).

Example 2:

Input: nums = [2,2,2]
Output: 1
Explanation: After including the reverse of each number, the resulting array is [2,2,2,2,2,2].
The number of distinct integers in this array is 1 (The number 2).

Constraints:

1 <= nums.length <= 10⁵
1 <= nums[i] <= 10⁶

Step 01

Brute Force Baseline

Problem summary: You are given an array nums consisting of positive integers. You have to take each integer in the array, reverse its digits, and add it to the end of the array. You should apply this operation to the original integers in nums. Return the number of distinct integers in the final array.

Baseline thinking

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

Pattern signal: Array · Hash Map · Math

Example 1

[1,13,10,12,31]

Example 2

[2,2,2]

Core Insight

What unlocks the optimal approach

What data structure allows us to insert numbers and find the number of distinct numbers in it?
Try using a set, insert all the numbers and their reverse into it, and return its size.

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 #2442: Count Number of Distinct Integers After Reverse Operations
class Solution {
    public int countDistinctIntegers(int[] nums) {
        Set<Integer> s = new HashSet<>();
        for (int x : nums) {
            s.add(x);
        }
        for (int x : nums) {
            int y = 0;
            while (x > 0) {
                y = y * 10 + x % 10;
                x /= 10;
            }
            s.add(y);
        }
        return s.size();
    }
}

// Accepted solution for LeetCode #2442: Count Number of Distinct Integers After Reverse Operations
func countDistinctIntegers(nums []int) int {
	s := map[int]struct{}{}
	for _, x := range nums {
		s[x] = struct{}{}
	}
	for _, x := range nums {
		y := 0
		for x > 0 {
			y = y*10 + x%10
			x /= 10
		}
		s[y] = struct{}{}
	}
	return len(s)
}

# Accepted solution for LeetCode #2442: Count Number of Distinct Integers After Reverse Operations
class Solution:
    def countDistinctIntegers(self, nums: List[int]) -> int:
        s = set(nums)
        for x in nums:
            y = int(str(x)[::-1])
            s.add(y)
        return len(s)

// Accepted solution for LeetCode #2442: Count Number of Distinct Integers After Reverse Operations
use std::collections::HashSet;
impl Solution {
    pub fn count_distinct_integers(nums: Vec<i32>) -> i32 {
        let mut set = HashSet::new();
        for i in 0..nums.len() {
            let mut num = nums[i];
            set.insert(num);
            set.insert({
                let mut item = 0;
                while num > 0 {
                    item = item * 10 + (num % 10);
                    num /= 10;
                }
                item
            });
        }
        set.len() as i32
    }
}

// Accepted solution for LeetCode #2442: Count Number of Distinct Integers After Reverse Operations
function countDistinctIntegers(nums: number[]): number {
    const n = nums.length;
    for (let i = 0; i < n; i++) {
        nums.push(Number([...(nums[i] + '')].reverse().join('')));
    }
    return new Set(nums).size;
}

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(n)

Approach Breakdown

BRUTE FORCE

O(n²) time

O(1) space

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.

OPTIMIZED

O(n) time

O(1) space

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.

Shortcut: If you are using nested loops on an array, there is almost always an O(n) solution. Look for the right auxiliary state.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

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.

Mutating counts without cleanup

Wrong move: Zero-count keys stay in map and break distinct/count constraints.

Usually fails on: Window/map size checks are consistently off by one.

Fix: Delete keys when count reaches zero.

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.