LeetCode #1814 — MEDIUM

Count Nice Pairs in an Array

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

The Problem

Problem Statement

You are given an array nums that consists of non-negative integers. Let us define rev(x) as the reverse of the non-negative integer x. For example, rev(123) = 321, and rev(120) = 21. A pair of indices (i, j) is nice if it satisfies all of the following conditions:

0 <= i < j < nums.length
nums[i] + rev(nums[j]) == nums[j] + rev(nums[i])

Return the number of nice pairs of indices. Since that number can be too large, return it modulo 10⁹ + 7.

Example 1:

Input: nums = [42,11,1,97]
Output: 2
Explanation: The two pairs are:
 - (0,3) : 42 + rev(97) = 42 + 79 = 121, 97 + rev(42) = 97 + 24 = 121.
 - (1,2) : 11 + rev(1) = 11 + 1 = 12, 1 + rev(11) = 1 + 11 = 12.

Example 2:

Input: nums = [13,10,35,24,76]
Output: 4

Constraints:

1 <= nums.length <= 10⁵
0 <= nums[i] <= 10⁹

Step 01

Brute Force Baseline

Problem summary: You are given an array nums that consists of non-negative integers. Let us define rev(x) as the reverse of the non-negative integer x. For example, rev(123) = 321, and rev(120) = 21. A pair of indices (i, j) is nice if it satisfies all of the following conditions: 0 <= i < j < nums.length nums[i] + rev(nums[j]) == nums[j] + rev(nums[i]) Return the number of nice pairs of indices. Since that number can be too large, return it modulo 109 + 7.

Baseline thinking

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

Pattern signal: Array · Hash Map · Math

Example 1

[42,11,1,97]

Example 2

[13,10,35,24,76]

Core Insight

What unlocks the optimal approach

The condition can be rearranged to (nums[i] - rev(nums[i])) == (nums[j] - rev(nums[j])).
Transform each nums[i] into (nums[i] - rev(nums[i])). Then, count the number of (i, j) pairs that have equal values.
Keep a map storing the frequencies of values that you have seen so far. For each i, check if nums[i] is in the map. If it is, then add that count to the overall count. Then, increment the frequency of nums[i].

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 #1814: Count Nice Pairs in an Array
class Solution {
    public int countNicePairs(int[] nums) {
        Map<Integer, Integer> cnt = new HashMap<>();
        for (int x : nums) {
            int y = x - rev(x);
            cnt.merge(y, 1, Integer::sum);
        }
        final int mod = (int) 1e9 + 7;
        long ans = 0;
        for (int v : cnt.values()) {
            ans = (ans + (long) v * (v - 1) / 2) % mod;
        }
        return (int) ans;
    }

    private int rev(int x) {
        int y = 0;
        for (; x > 0; x /= 10) {
            y = y * 10 + x % 10;
        }
        return y;
    }
}

// Accepted solution for LeetCode #1814: Count Nice Pairs in an Array
func countNicePairs(nums []int) (ans int) {
	rev := func(x int) (y int) {
		for ; x > 0; x /= 10 {
			y = y*10 + x%10
		}
		return
	}
	cnt := map[int]int{}
	for _, x := range nums {
		y := x - rev(x)
		cnt[y]++
	}
	const mod int = 1e9 + 7
	for _, v := range cnt {
		ans = (ans + v*(v-1)/2) % mod
	}
	return
}

# Accepted solution for LeetCode #1814: Count Nice Pairs in an Array
class Solution:
    def countNicePairs(self, nums: List[int]) -> int:
        def rev(x):
            y = 0
            while x:
                y = y * 10 + x % 10
                x //= 10
            return y

        cnt = Counter(x - rev(x) for x in nums)
        mod = 10**9 + 7
        return sum(v * (v - 1) // 2 for v in cnt.values()) % mod

// Accepted solution for LeetCode #1814: Count Nice Pairs in an Array
struct Solution;

use std::collections::HashMap;

const MOD: i64 = 1_000_000_007;

impl Solution {
    fn count_nice_pairs(nums: Vec<i32>) -> i32 {
        let mut hm: HashMap<i32, i32> = HashMap::new();
        let mut res: i64 = 0;
        for x in nums {
            let diff = x - rev(x);
            let count = hm.entry(diff).or_default();
            res += *count as i64;
            *count += 1;
        }
        (res % MOD) as i32
    }
}

fn rev(mut x: i32) -> i32 {
    let mut digits: Vec<i32> = vec![];
    while x > 0 {
        digits.push(x % 10);
        x /= 10;
    }
    let mut res = 0;
    for x in digits {
        res *= 10;
        res += x;
    }
    res
}

#[test]
fn test() {
    let nums = vec![42, 11, 1, 97];
    let res = 2;
    assert_eq!(Solution::count_nice_pairs(nums), res);
    let nums = vec![13, 10, 35, 24, 76];
    let res = 4;
    assert_eq!(Solution::count_nice_pairs(nums), res);
}

// Accepted solution for LeetCode #1814: Count Nice Pairs in an Array
function countNicePairs(nums: number[]): number {
    const rev = (x: number): number => {
        let y = 0;
        while (x) {
            y = y * 10 + (x % 10);
            x = Math.floor(x / 10);
        }
        return y;
    };
    const mod = 10 ** 9 + 7;
    const cnt = new Map<number, number>();
    let ans = 0;
    for (const x of nums) {
        const y = x - rev(x);
        ans = (ans + (cnt.get(y) ?? 0)) % mod;
        cnt.set(y, (cnt.get(y) ?? 0) + 1);
    }
    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(n)

Space

O(1)

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.