Count of substrings of length K with exactly K-1 distinct characters
Last Updated :
28 Mar, 2025
Given a string consisting of lowercase characters and an integer k, the task is to count all substrings of length k which have exactly k-1 distinct characters.
Example:
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.
Input: s = "aabab", k = 3
Output: 3
Explanation: Substrings of length 3 are "aab", "aba", "bab". All 2 substrings contains k-1 distinct characters.
[Naive Approach] Checking each Substring - O(n * k) time and O(1) space
The idea is to consider each substring of length k, and check if it contains exactly k-1 distinct characters.
C++
#include <bits/stdc++.h>
using namespace std;
int countOfSubstrings(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 str = "abcc";
int k = 2;
cout << countOfSubstrings(str, k) << endl;
return 0;
}
class GfG {
static int countOfSubstrings(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 str = "abcc";
int k = 2;
System.out.println(countOfSubstrings(str, k));
}
}
def countOfSubstrings(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
if __name__ == "__main__":
str = "abcc"
k = 2
print(countOfSubstrings(str, k))
using System;
class GfG {
static int countOfSubstrings(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 str = "abcc";
int k = 2;
Console.WriteLine(countOfSubstrings(str, k));
}
}
function countOfSubstrings(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;
}
let str = "abcc";
let k = 2;
console.log(countOfSubstrings(str, k));
[Expected Approach] Using Sliding Window Method - O(n) time and O(1) space
The idea is to use a sliding window of length k across the string and count how many windows contain exactly k-1 distinct characters. We maintain a frequency counter for each character in the current window and check if the number of distinct characters (characters with non-zero frequency) is equal to k-1.
Step by step approach:
- Initialize a frequency array to track the count of each lowercase character in the current window.
- Build the initial window of size k-1, then slide through the string one character at a time.
- For each window position, add the new character and check if exactly k-1 distinct characters exist. If the condition is met, increment the result counter.
- Before sliding the window, remove the leftmost character from the frequency count.
C++
#include <bits/stdc++.h>
using namespace std;
int countOfSubstrings(string &s, int k) {
// If k is greater than string length
if (k > s.length()) return 0;
int n = s.length(), j = 0;
vector<int> cnt(26, 0);
int ans = 0;
for (int i=0; i<k-1; i++) {
cnt[s[i]-'a']++;
}
for (int i=k-1; i<n; i++) {
cnt[s[i]-'a']++;
// Check if the current window
// contains k-1 distinct chars.
int distinctCnt = 0;
for (int j=0; j<26; j++) {
if (cnt[j] > 0) distinctCnt++;
}
if (distinctCnt == k-1) ans++;
cnt[s[i-k+1]-'a']--;
}
return ans;
}
int main() {
string str = "abcc";
int k = 2;
cout << countOfSubstrings(str, k) << endl;
return 0;
}
class GfG {
static int countOfSubstrings(String s, int k) {
if (k > s.length()) return 0;
int n = s.length();
int[] cnt = new int[26];
int ans = 0;
for (int i = 0; i < k - 1; i++) {
cnt[s.charAt(i) - 'a']++;
}
for (int i = k - 1; i < n; i++) {
cnt[s.charAt(i) - 'a']++;
// Check if the current window
// contains k-1 distinct chars.
int distinctCnt = 0;
for (int j1 = 0; j1 < 26; j1++) {
if (cnt[j1] > 0) distinctCnt++;
}
if (distinctCnt == k - 1) ans++;
cnt[s.charAt(i - k + 1) - 'a']--;
}
return ans;
}
public static void main(String[] args) {
String str = "abcc";
int k = 2;
System.out.println(countOfSubstrings(str, k));
}
}
def countOfSubstrings(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__":
str = "abcc"
k = 2
print(countOfSubstrings(str, k))
using System;
class GfG {
static int countOfSubstrings(string s, int k) {
if (k > s.Length) return 0;
int n = s.Length;
int[] cnt = new int[26];
int ans = 0;
for (int i = 0; i < k - 1; i++) {
cnt[s[i] - 'a']++;
}
for (int i = k - 1; i < n; i++) {
cnt[s[i] - 'a']++;
// Check if the current window
// contains k-1 distinct chars.
int distinctCnt = 0;
for (int j1 = 0; j1 < 26; j1++) {
if (cnt[j1] > 0) distinctCnt++;
}
if (distinctCnt == k - 1) ans++;
cnt[s[i - k + 1] - 'a']--;
}
return ans;
}
static void Main() {
string str = "abcc";
int k = 2;
Console.WriteLine(countOfSubstrings(str, k));
}
}
function countOfSubstrings(s, k) {
// If k is greater than string length
if (k > s.length) return 0;
let n = s.length;
let cnt = new Array(26).fill(0);
let ans = 0;
for (let i = 0; i < k - 1; i++) {
cnt[s.charCodeAt(i) - 'a'.charCodeAt(0)]++;
}
for (let i = k - 1; i < n; i++) {
cnt[s.charCodeAt(i) - 'a'.charCodeAt(0)]++;
// Check if the current window
// contains k-1 distinct chars.
let distinctCnt = cnt.filter(x => x > 0).length;
if (distinctCnt === k - 1) ans++;
cnt[s.charCodeAt(i - k + 1) - 'a'.charCodeAt(0)]--;
}
return ans;
}
let str = "abcc";
let k = 2;
console.log(countOfSubstrings(str, k));
Similar Reads
Count substrings with k distinct characters Given a string s consisting of lowercase characters and an integer k, the task is to count all possible substrings (not necessarily distinct) that have exactly k distinct characters. Examples: Input: s = "abc", k = 2Output: 2Explanation: Possible substrings are ["ab", "bc"]Input: s = "aba", k = 2Out
10 min read
Number of substrings with count of each character as k Given a string and an integer k, find the number of substrings in which all the different characters occur exactly k times. Examples: Input : s = "aabbcc" k = 2 Output : 6 The substrings are aa, bb, cc, aabb, bbcc and aabbcc. Input : s = "aabccc" k = 2 Output : 3 There are three substrings aa, cc an
15 min read
Count of substrings having all distinct characters Given a string str consisting of lowercase alphabets, the task is to find the number of possible substrings (not necessarily distinct) that consists of distinct characters only.Examples: Input: Str = "gffg" Output: 6 Explanation: All possible substrings from the given string are, ( "g", "gf", "gff",
7 min read
Count number of substrings having at least K distinct characters Given a string S consisting of N characters and a positive integer K, the task is to count the number of substrings having at least K distinct characters.Examples:Input: S = "abcca", K = 3Output: 4Explanation:The substrings that contain at least K(= 3) distinct characters are:"abc": Count of distinc
7 min read
Count of Substrings with at least K pairwise Distinct Characters having same Frequency Given a string S and an integer K, the task is to find the number of substrings which consists of at least K pairwise distinct characters having same frequency. Examples: Input: S = "abasa", K = 2 Output: 5 Explanation: The substrings in having 2 pairwise distinct characters with same frequency are
7 min read