LeetCode #1915 — MEDIUM

Number of Wonderful Substrings

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

The Problem

Problem Statement

A wonderful string is a string where at most one letter appears an odd number of times.

For example, "ccjjc" and "abab" are wonderful, but "ab" is not.

Given a string word that consists of the first ten lowercase English letters ('a' through 'j'), return the number of wonderful non-empty substrings in word. If the same substring appears multiple times in word, then count each occurrence separately.

A substring is a contiguous sequence of characters in a string.

Example 1:

Input: word = "aba"
Output: 4
Explanation: The four wonderful substrings are underlined below:
- "aba" -> "a"
- "aba" -> "b"
- "aba" -> "a"
- "aba" -> "aba"

Example 2:

Input: word = "aabb"
Output: 9
Explanation: The nine wonderful substrings are underlined below:
- "aabb" -> "a"
- "aabb" -> "aa"
- "aabb" -> "aab"
- "aabb" -> "aabb"
- "aabb" -> "a"
- "aabb" -> "abb"
- "aabb" -> "b"
- "aabb" -> "bb"
- "aabb" -> "b"

Example 3:

Input: word = "he"
Output: 2
Explanation: The two wonderful substrings are underlined below:
- "he" -> "h"
- "he" -> "e"

Constraints:

1 <= word.length <= 10⁵
word consists of lowercase English letters from 'a' to 'j'.

Step 01

Brute Force Baseline

Problem summary: A wonderful string is a string where at most one letter appears an odd number of times. For example, "ccjjc" and "abab" are wonderful, but "ab" is not. Given a string word that consists of the first ten lowercase English letters ('a' through 'j'), return the number of wonderful non-empty substrings in word. If the same substring appears multiple times in word, then count each occurrence separately. A substring is a contiguous sequence of characters in a string.

Baseline thinking

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

Pattern signal: Hash Map · Bit Manipulation

Example 1

"aba"

Example 2

"aabb"

Example 3

"he"

Step 02

Core Insight

What unlocks the optimal approach

For each prefix of the string, check which characters are of even frequency and which are not and represent it by a bitmask.
Find the other prefixes whose masks differs from the current prefix mask by at most one bit.

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 #1915: Number of Wonderful Substrings
class Solution {
    public long wonderfulSubstrings(String word) {
        int[] cnt = new int[1 << 10];
        cnt[0] = 1;
        long ans = 0;
        int st = 0;
        for (char c : word.toCharArray()) {
            st ^= 1 << (c - 'a');
            ans += cnt[st];
            for (int i = 0; i < 10; ++i) {
                ans += cnt[st ^ (1 << i)];
            }
            ++cnt[st];
        }
        return ans;
    }
}

// Accepted solution for LeetCode #1915: Number of Wonderful Substrings
func wonderfulSubstrings(word string) (ans int64) {
	cnt := [1024]int{1}
	st := 0
	for _, c := range word {
		st ^= 1 << (c - 'a')
		ans += int64(cnt[st])
		for i := 0; i < 10; i++ {
			ans += int64(cnt[st^(1<<i)])
		}
		cnt[st]++
	}
	return
}

# Accepted solution for LeetCode #1915: Number of Wonderful Substrings
class Solution:
    def wonderfulSubstrings(self, word: str) -> int:
        cnt = Counter({0: 1})
        ans = st = 0
        for c in word:
            st ^= 1 << (ord(c) - ord("a"))
            ans += cnt[st]
            for i in range(10):
                ans += cnt[st ^ (1 << i)]
            cnt[st] += 1
        return ans

// Accepted solution for LeetCode #1915: Number of Wonderful Substrings
/**
 * [1915] Number of Wonderful Substrings
 *
 * A wonderful string is a string where at most one letter appears an odd number of times.
 *
 *
 * 	For example, "ccjjc" and "abab" are wonderful, but "ab" is not.
 *
 *
 * Given a string word that consists of the first ten lowercase English letters ('a' through 'j'), return the number of wonderful non-empty substrings in word. If the same substring appears multiple times in word, then count each occurrence separately.
 *
 * A substring is a contiguous sequence of characters in a string.
 *
 *  
 * Example 1:
 *
 *
 * Input: word = "aba"
 * Output: 4
 * Explanation: The four wonderful substrings are underlined below:
 * - "<u>a</u>ba" -> "a"
 * - "a<u>b</u>a" -> "b"
 * - "ab<u>a</u>" -> "a"
 * - "<u>aba</u>" -> "aba"
 *
 *
 * Example 2:
 *
 *
 * Input: word = "aabb"
 * Output: 9
 * Explanation: The nine wonderful substrings are underlined below:
 * - "<u>a</u>abb" -> "a"
 * - "<u>aa</u>bb" -> "aa"
 * - "<u>aab</u>b" -> "aab"
 * - "<u>aabb</u>" -> "aabb"
 * - "a<u>a</u>bb" -> "a"
 * - "a<u>abb</u>" -> "abb"
 * - "aa<u>b</u>b" -> "b"
 * - "aa<u>bb</u>" -> "bb"
 * - "aab<u>b</u>" -> "b"
 *
 *
 * Example 3:
 *
 *
 * Input: word = "he"
 * Output: 2
 * Explanation: The two wonderful substrings are underlined below:
 * - "<u>h</u>e" -> "h"
 * - "h<u>e</u>" -> "e"
 *
 *
 *  
 * Constraints:
 *
 *
 * 	1 <= word.length <= 10^5
 * 	word consists of lowercase English letters from 'a' to 'j'.
 *
 */
pub struct Solution {}

// problem: https://leetcode.com/problems/number-of-wonderful-substrings/
// discuss: https://leetcode.com/problems/number-of-wonderful-substrings/discuss/?currentPage=1&orderBy=most_votes&query=

// submission codes start here

impl Solution {
    pub fn wonderful_substrings(word: String) -> i64 {
        word.bytes()
            .fold(
                (0usize, 0i64, {
                    let mut v = vec![0i64; 1024];
                    v[0] = 1;
                    v
                }),
                |(mut x, mut ans, mut v), c| {
                    x ^= 1 << (c - 97);
                    ans += v[x];
                    ans += (0..10).map(|i| v[x ^ (1 << i)]).sum::<i64>();
                    v[x] += 1;
                    (x, ans, v)
                },
            )
            .1
    }
}

// submission codes end

#[cfg(test)]
mod tests {
    use super::*;

    #[test]
    fn test_1915_example_1() {
        let word = "aba".to_string();

        let result = 4;

        assert_eq!(Solution::wonderful_substrings(word), result);
    }

    #[test]
    fn test_1915_example_2() {
        let word = "aabb".to_string();

        let result = 9;

        assert_eq!(Solution::wonderful_substrings(word), result);
    }

    #[test]
    fn test_1915_example_3() {
        let word = "he".to_string();

        let result = 2;

        assert_eq!(Solution::wonderful_substrings(word), result);
    }
}

// Accepted solution for LeetCode #1915: Number of Wonderful Substrings
function wonderfulSubstrings(word: string): number {
    const cnt: number[] = new Array(1 << 10).fill(0);
    cnt[0] = 1;
    let ans = 0;
    let st = 0;
    for (const c of word) {
        st ^= 1 << (c.charCodeAt(0) - 'a'.charCodeAt(0));
        ans += cnt[st];
        for (let i = 0; i < 10; ++i) {
            ans += cnt[st ^ (1 << i)];
        }
        cnt[st]++;
    }
    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

SORT + SCAN

O(n log n) time

O(n) space

Sort the array in O(n log n), then scan for the missing or unique element by comparing adjacent pairs. Sorting requires O(n) auxiliary space (or O(1) with in-place sort but O(n log n) time remains). The sort step dominates.

BIT MANIPULATION

O(n) time

O(1) space

Bitwise operations (AND, OR, XOR, shifts) are O(1) per operation on fixed-width integers. A single pass through the input with bit operations gives O(n) time. The key insight: XOR of a number with itself is 0, which eliminates duplicates without extra space.

Shortcut: Bit operations are O(1). XOR cancels duplicates. Single pass → O(n) time, O(1) space.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

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.