LeetCode #3267 — HARD

Count Almost Equal Pairs II

Break down a hard problem into reliable checkpoints, edge-case handling, and complexity trade-offs.

The Problem

Problem Statement

Attention: In this version, the number of operations that can be performed, has been increased to twice.

You are given an array nums consisting of positive integers.

We call two integers x and y almost equal if both integers can become equal after performing the following operation at most twice:

Choose either x or y and swap any two digits within the chosen number.

Return the number of indices i and j in nums where i < j such that nums[i] and nums[j] are almost equal.

Note that it is allowed for an integer to have leading zeros after performing an operation.

Example 1:

Input: nums = [1023,2310,2130,213]

Output: 4

Explanation:

The almost equal pairs of elements are:

1023 and 2310. By swapping the digits 1 and 2, and then the digits 0 and 3 in 1023, you get 2310.
1023 and 213. By swapping the digits 1 and 0, and then the digits 1 and 2 in 1023, you get 0213, which is 213.
2310 and 213. By swapping the digits 2 and 0, and then the digits 3 and 2 in 2310, you get 0213, which is 213.
2310 and 2130. By swapping the digits 3 and 1 in 2310, you get 2130.

Example 2:

Input: nums = [1,10,100]

Output: 3

Explanation:

The almost equal pairs of elements are:

1 and 10. By swapping the digits 1 and 0 in 10, you get 01 which is 1.
1 and 100. By swapping the second 0 with the digit 1 in 100, you get 001, which is 1.
10 and 100. By swapping the first 0 with the digit 1 in 100, you get 010, which is 10.

Constraints:

2 <= nums.length <= 5000
1 <= nums[i] < 10⁷

Step 01

Brute Force Baseline

Problem summary: Attention: In this version, the number of operations that can be performed, has been increased to twice. You are given an array nums consisting of positive integers. We call two integers x and y almost equal if both integers can become equal after performing the following operation at most twice: Choose either x or y and swap any two digits within the chosen number. Return the number of indices i and j in nums where i < j such that nums[i] and nums[j] are almost equal. Note that it is allowed for an integer to have leading zeros after performing an operation.

Baseline thinking

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

Pattern signal: Array · Hash Map

Example 1

[1023,2310,2130,213]

Example 2

[1,10,100]

Core Insight

What unlocks the optimal approach

For each element, find all possible integers we can get by applying the operations.
Store the frequencies of all the integers in a hashmap.

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

Largest constraint values

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 #3267: Count Almost Equal Pairs II
class Solution {
    public int countPairs(int[] nums) {
        Arrays.sort(nums);
        int ans = 0;
        Map<Integer, Integer> cnt = new HashMap<>();
        for (int x : nums) {
            Set<Integer> vis = new HashSet<>();
            vis.add(x);
            char[] s = String.valueOf(x).toCharArray();
            for (int j = 0; j < s.length; ++j) {
                for (int i = 0; i < j; ++i) {
                    swap(s, i, j);
                    vis.add(Integer.parseInt(String.valueOf(s)));
                    for (int q = i; q < s.length; ++q) {
                        for (int p = i; p < q; ++p) {
                            swap(s, p, q);
                            vis.add(Integer.parseInt(String.valueOf(s)));
                            swap(s, p, q);
                        }
                    }
                    swap(s, i, j);
                }
            }
            for (int y : vis) {
                ans += cnt.getOrDefault(y, 0);
            }
            cnt.merge(x, 1, Integer::sum);
        }
        return ans;
    }

    private void swap(char[] s, int i, int j) {
        char t = s[i];
        s[i] = s[j];
        s[j] = t;
    }
}

// Accepted solution for LeetCode #3267: Count Almost Equal Pairs II
func countPairs(nums []int) (ans int) {
	sort.Ints(nums)
	cnt := make(map[int]int)

	for _, x := range nums {
		vis := make(map[int]struct{})
		vis[x] = struct{}{}
		s := []rune(strconv.Itoa(x))

		for j := 0; j < len(s); j++ {
			for i := 0; i < j; i++ {
				s[i], s[j] = s[j], s[i]
				y, _ := strconv.Atoi(string(s))
				vis[y] = struct{}{}
				for q := i + 1; q < len(s); q++ {
					for p := i + 1; p < q; p++ {
						s[p], s[q] = s[q], s[p]
						z, _ := strconv.Atoi(string(s))
						vis[z] = struct{}{}
						s[p], s[q] = s[q], s[p]
					}
				}
				s[i], s[j] = s[j], s[i]
			}
		}

		for y := range vis {
			ans += cnt[y]
		}
		cnt[x]++
	}

	return
}

# Accepted solution for LeetCode #3267: Count Almost Equal Pairs II
class Solution:
    def countPairs(self, nums: List[int]) -> int:
        nums.sort()
        ans = 0
        cnt = defaultdict(int)
        for x in nums:
            vis = {x}
            s = list(str(x))
            m = len(s)
            for j in range(m):
                for i in range(j):
                    s[i], s[j] = s[j], s[i]
                    vis.add(int("".join(s)))
                    for q in range(i + 1, m):
                        for p in range(i + 1, q):
                            s[p], s[q] = s[q], s[p]
                            vis.add(int("".join(s)))
                            s[p], s[q] = s[q], s[p]
                    s[i], s[j] = s[j], s[i]
            ans += sum(cnt[x] for x in vis)
            cnt[x] += 1
        return ans

// Accepted solution for LeetCode #3267: Count Almost Equal Pairs II
// Rust example auto-generated from java reference.
// Replace the signature and local types with the exact LeetCode harness for this problem.
impl Solution {
    pub fn rust_example() {
        // Port the logic from the reference block below.
    }
}

// Reference (java):
// // Accepted solution for LeetCode #3267: Count Almost Equal Pairs II
// class Solution {
//     public int countPairs(int[] nums) {
//         Arrays.sort(nums);
//         int ans = 0;
//         Map<Integer, Integer> cnt = new HashMap<>();
//         for (int x : nums) {
//             Set<Integer> vis = new HashSet<>();
//             vis.add(x);
//             char[] s = String.valueOf(x).toCharArray();
//             for (int j = 0; j < s.length; ++j) {
//                 for (int i = 0; i < j; ++i) {
//                     swap(s, i, j);
//                     vis.add(Integer.parseInt(String.valueOf(s)));
//                     for (int q = i; q < s.length; ++q) {
//                         for (int p = i; p < q; ++p) {
//                             swap(s, p, q);
//                             vis.add(Integer.parseInt(String.valueOf(s)));
//                             swap(s, p, q);
//                         }
//                     }
//                     swap(s, i, j);
//                 }
//             }
//             for (int y : vis) {
//                 ans += cnt.getOrDefault(y, 0);
//             }
//             cnt.merge(x, 1, Integer::sum);
//         }
//         return ans;
//     }
// 
//     private void swap(char[] s, int i, int j) {
//         char t = s[i];
//         s[i] = s[j];
//         s[j] = t;
//     }
// }

// Accepted solution for LeetCode #3267: Count Almost Equal Pairs II
function countPairs(nums: number[]): number {
    nums.sort((a, b) => a - b);
    let ans = 0;
    const cnt = new Map<number, number>();

    for (const x of nums) {
        const vis = new Set<number>();
        vis.add(x);
        const s = x.toString().split('');

        for (let j = 0; j < s.length; j++) {
            for (let i = 0; i < j; i++) {
                [s[i], s[j]] = [s[j], s[i]];
                vis.add(+s.join(''));
                for (let q = i + 1; q < s.length; ++q) {
                    for (let p = i + 1; p < q; ++p) {
                        [s[p], s[q]] = [s[q], s[p]];
                        vis.add(+s.join(''));
                        [s[p], s[q]] = [s[q], s[p]];
                    }
                }
                [s[i], s[j]] = [s[j], s[i]];
            }
        }

        for (const y of vis) {
            ans += cnt.get(y) || 0;
        }
        cnt.set(x, (cnt.get(x) || 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 × (log n + log^5 M)

Space

O(n + log^4 M)

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.