Count of substrings of length K with exactly K-1 distinct characters
Last Updated :
01 Jul, 2025
Given a string s consisting of lowercase characters and an integer k, find the count of all substrings of length k which have exactly k-1 distinct characters.
Example:
Input: s = "aabab", k = 3
Output: 3
Explanation: Substrings of length 3 are "aab", "aba", "bab". All 3 substrings contains 2 distinct characters.
Input: s = "abcc", k = 2
Output: 1
Explanation: Substrings of length 2 are "ab", "bc" and "cc". Only "cc" has 2-1 = 1 distinct character.
[Naive Approach] Checking each substring- O(n * k) time and O(1) space
The main idea of this approach is to iterate through all possible substrings of length k in the given string s. For each substring, it uses a frequency array to count the occurrences of each character. Then, it calculates how many distinct characters are present in that substring. If the number of distinct characters equals k - 1, the substring satisfies the condition and the count is incremented.
C++
#include <bits/stdc++.h>
using namespace std;
int substrCount(string &s, int k) {
int n = s.length();
int count = 0;
// Check for all subarray of size K
for (int i = 0; i < n - k + 1; i++) {
vector<int> cnt(26, 0);
for (int j=i; j<i+k; j++)
cnt[s[j] - 'a']++;
int distinctCnt = 0;
// Count number of distinct characters
for (int i=0; i<26; i++) {
if (cnt[i] > 0) {
distinctCnt++;
}
}
if (distinctCnt == k-1) count++;
}
return count;
}
int main() {
string s = "aabab";
int k =3;
cout << substrCount(s, k) << endl;
return 0;
}
class GfG {
static int substrCount(String s, int k) {
int n = s.length();
int count = 0;
// Check for all subarray of size K
for (int i = 0; i < n - k + 1; i++) {
int[] cnt = new int[26];
for (int j = i; j < i + k; j++)
cnt[s.charAt(j) - 'a']++;
int distinctCnt = 0;
// Count number of distinct characters
for (int i1 = 0; i1 < 26; i1++) {
if (cnt[i1] > 0) {
distinctCnt++;
}
}
if (distinctCnt == k - 1) count++;
}
return count;
}
public static void main(String[] args) {
String s = "aabab";
int k = 3;
System.out.println(substrCount(s, k));
}
}
def substrCount(s, k):
n = len(s)
count = 0
# Check for all subarray of size K
for i in range(n - k + 1):
cnt = [0] * 26
for j in range(i, i + k):
cnt[ord(s[j]) - ord('a')] += 1
distinctCnt = sum(1 for x in cnt if x > 0)
if distinctCnt == k - 1:
count += 1
return count
# Driver Code
if __name__ == "__main__":
s = "aabab"
k = 3
print(substrCount(s, k))
using System;
class GfG {
static int substrCount(string s, int k) {
int n = s.Length;
int count = 0;
// Check for all subarray of size K
for (int i = 0; i < n - k + 1; i++) {
int[] cnt = new int[26];
for (int j = i; j < i + k; j++)
cnt[s[j] - 'a']++;
int distinctCnt = 0;
// Count number of distinct characters
for (int i1 = 0; i1 < 26; i1++) {
if (cnt[i1] > 0) {
distinctCnt++;
}
}
if (distinctCnt == k - 1) count++;
}
return count;
}
public static void Main(string[] args) {
string s = "aabab";
int k = 3;
Console.WriteLine(substrCount(s, k));
}
}
function substrCount(s, k) {
let n = s.length;
let count = 0;
// Check for all subarray of size K
for (let i = 0; i < n - k + 1; i++) {
let cnt = new Array(26).fill(0);
for (let j = i; j < i + k; j++)
cnt[s.charCodeAt(j) - 'a'.charCodeAt(0)]++;
let distinctCnt = cnt.filter(x => x > 0).length;
if (distinctCnt === k - 1) count++;
}
return count;
}
// Dricer Code
let s = "aabab";
let k = 3;
console.log(substrCount(s, k));
[Expected Approach] Using Sliding Window Method - O(n) time and O(1) space
The idea is to maintains a 26 element frequency array for letters a–z and a counter distinctCnt for how many different letters are currently in the window. First, it preloads the initial k - 1 characters so that each subsequent step merely adds one new character (expanding the window to size k), checks whether the window now contains exactly k - 1 distinct letters, and, if so, increments the answer. Then it removes the leftmost character to keep the window size constant before moving to the next index.
Illustration:
C++
#include <bits/stdc++.h>
using namespace std;
int substrCount(string &s, int k) {
// Return 0 if window size k is greater than the string length
if (k > s.length()) return 0;
int n = s.length();
// Frequency array for characters a-z
vector<int> cnt(26, 0);
int ans = 0;
int distinctCnt = 0;
// Preprocess the first k-1 characters
for (int i = 0; i < k - 1; i++) {
cnt[s[i] - 'a']++;
if (cnt[s[i] - 'a'] == 1) distinctCnt++;
}
// Start sliding window
for (int i = k - 1; i < n; i++) {
cnt[s[i] - 'a']++;
if (cnt[s[i] - 'a'] == 1) distinctCnt++;
// Check if window has exactly k-1 distinct characters
if (distinctCnt == k - 1) ans++;
// Remove the character going out of the window
cnt[s[i - k + 1] - 'a']--;
if (cnt[s[i - k + 1] - 'a'] == 0) distinctCnt--;
}
return ans;
}
int main() {
string str = "aabab";
int k = 3;
cout << substrCount(str, k) << endl;
return 0;
}
import java.util.*;
class GfG {
static int substrCount(String s, int k) {
// If k is larger than the string length, return 0
if (k > s.length()) return 0;
int n = s.length();
// Frequency array for characters 'a' to 'z'
int[] cnt = new int[26];
int ans = 0;
int distinctCnt = 0;
// Pre-fill the first (k-1) characters in frequency array
for (int i = 0; i < k - 1; i++) {
int index = s.charAt(i) - 'a';
cnt[index]++;
// New distinct character
if (cnt[index] == 1) distinctCnt++;
}
// Start sliding window of size k
for (int i = k - 1; i < n; i++) {
int addIndex = s.charAt(i) - 'a';
cnt[addIndex]++;
// New distinct character added
if (cnt[addIndex] == 1) distinctCnt++;
// If exactly (k - 1) distinct characters, it's valid
if (distinctCnt == k - 1) ans++;
// Remove the character going out from the left of window
int removeIndex = s.charAt(i - k + 1) - 'a';
cnt[removeIndex]--;
// One distinct character removed
if (cnt[removeIndex] == 0) distinctCnt--;
}
return ans;
}
public static void main(String[] args) {
String s = "aabab";
int k = 3;
System.out.println(substrCount(s, k));
}
}
def substrCount(s, k):
if k > len(s):
return 0
n = len(s)
cnt = [0] * 26
ans = 0
for i in range(k - 1):
cnt[ord(s[i]) - ord('a')] += 1
for i in range(k - 1, n):
cnt[ord(s[i]) - ord('a')] += 1
# Check if the current window
# contains k-1 distinct chars.
distinctCnt = sum(1 for x in cnt if x > 0)
if distinctCnt == k - 1:
ans += 1
cnt[ord(s[i - k + 1]) - ord('a')] -= 1
return ans
if __name__ == "__main__":
s = "aabab"
k = 3
print(substrCount(s, k))
using System;
class GfG{
static int substrCount(string s, int k){
// If k is greater than string length, return 0
if (k > s.Length) return 0;
int n = s.Length;
// Frequency array for a-z
int[] cnt = new int[26];
int ans = 0;
int distinctCnt = 0;
// Preprocess first (k - 1) characters
for (int i = 0; i < k - 1; i++){
int index = s[i] - 'a';
cnt[index]++;
if (cnt[index] == 1) distinctCnt++;
}
// Slide the window across the string
for (int i = k - 1; i < n; i++){
int addIndex = s[i] - 'a';
cnt[addIndex]++;
if (cnt[addIndex] == 1) distinctCnt++;
// Check if window has exactly k - 1 distinct characters
if (distinctCnt == k - 1) ans++;
// Remove character from the left of the window
int removeIndex = s[i - k + 1] - 'a';
cnt[removeIndex]--;
if (cnt[removeIndex] == 0) distinctCnt--;
}
return ans;
}
static void Main(){
string s = "aabab";
int k = 3;
Console.WriteLine(substrCount(s, k));
}
}
function substrCount(s, k) {
// Edge case: window larger than string
if (k > s.length) return 0;
const n = s.length;
// frequency array for 'a' … 'z'
const cnt = new Array(26).fill(0);
let ans = 0;
let distinctCnt = 0;
// Preload first k-1 characters
for (let i = 0; i < k - 1; i++) {
const idx = s.charCodeAt(i) - 97;
// new distinct char
if (++cnt[idx] === 1) distinctCnt++;
}
// Slide window of exact size k
for (let i = k - 1; i < n; i++) {
const addIdx = s.charCodeAt(i) - 97;
if (++cnt[addIdx] === 1) distinctCnt++;
if (distinctCnt === k - 1) ans++;
const remIdx = s.charCodeAt(i - k + 1) - 97;
if (--cnt[remIdx] === 0) distinctCnt--;
}
return ans;
}
// Driver Code
let s = "aabab";
let k = 3;
console.log(substrCount(s, k));
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