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Count of substrings with the frequency of at most one character as Odd

  • Difficulty Level : Medium
  • Last Updated : 01 Oct, 2021

Given a string S of N characters, the task is to calculate the total number of non-empty substrings such that at most one character occurs an odd number of times.


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Input: S = “aba”
Output: 4
Explanation: The valid substrings are “a”, “b”, “a”, and “aba”. Therefore, the total number of required substrings are 4.

Input: “aabb”
Output: 9
Explanation: The valid substrings are “a”, “aa”, “aab”, “aabb”, “a”, “abb”, “b”, “bb”, and “b”.

Approach: The above problem can be solved with the help of Bit Masking using HashMaps. Follow the below-mentioned steps to solve the problem:

  • The parity of the frequency of each character can be stored in a bitmask mask, where the ith character is represented by 2i. Initially the value of mask = 0. 
  • Create an unordered map seen, which stores the frequency of occurrence of each bitmask. Initially, the value of  seen[0] = 1.
  • Create a variable cnt, which stores the count of the valid substrings. Initially, the value of cnt = 0.
  • Iterate for each i in the range [0, N) and Bitwise XOR the value of the mask with the integer representing the ith character of the string and increment the value of cnt by seen[mask].
  • For each valid i, Iterate through all characters in the range [a, z] and increase its frequency by flipping the jth set-bit in the current mask and increment the value of the cnt by the frequency of bitmask after flipping the jth set-bit.
  • The value stored in cnt is the required answer.

Below is the implementation of the above approach:


// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
// Function to find the count of substrings
// such that at most one character occurs
// odd number of times
int uniqueSubstrings(string S)
    // Stores the frequency of the bitmasks
    unordered_map<int, int> seen;
    // Initial Condition
    seen[0] = 1;
    // Store the current value of the bitmask
    int mask = 0;
    // Stores the total count of the
    // valid substrings
    int cnt = 0;
    for (int i = 0; i < S.length(); ++i) {
        // XOR the mask with current character
        mask ^= (1 << (S[i] - 'a'));
        // Increment the count by mask count
        // of strings with all even frequencies
        cnt += seen[mask];
        for (int j = 0; j < 26; ++j) {
            // Increment count by mask count
            // of strings if exist with the
            // jth character having odd frequency
            cnt += seen[mask ^ (1 << j)];
    // Return Answer
    return cnt;
// Driver Code
int main()
    string word = "aabb";
    cout << uniqueSubstrings(word);
    return 0;



Time Complexity: O(N)
Auxiliary Space: O(N)

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