Smallest window that contains all characters of string itself
Last Updated :
23 Jul, 2025
Given a string str, your task is to find the smallest window length that contains all the characters of the given string at least one time.
Examples:
Input: str = "aabcbcdbca"
Output: 4
Explanation: Sub-string -> "dbca"
Input: str = "aaab"
Output: 2
Explanation: Sub-string -> "ab"
[Naive Approach] - Generating All Substrings - O(n ^ 2) Time and O(1) Space
The idea is to first identify all the unique characters using a count array. Then, by generating all possible substrings and checking whether each contains all the unique characters, we can track and update the shortest valid substring found. The approach involves efficiently tracking which characters have been seen using a visited array and leveraging two pointers to examine different substrings. Once a substring contains all distinct characters, we compare its length with the minimum length found so far and update the result if necessary.
C++
#include <bits/stdc++.h>
using namespace std;
int findSubString(string &str) {
int n = str.size();
int ans = n;
// to store all distinct characters
vector<bool> visited(26, false);
for (int i = 0; i < n; i++) {
visited[str[i]-'a'] = true;
}
for(int i = 0; i < n; i++) {
vector<bool> cur(26, false);
for(int j = i; j < n; j++) {
cur[str[j]-'a'] = true;
// if all characters are present
if(cur == visited) {
ans = min(ans, j - i + 1);
break;
}
}
}
return ans;
}
int main() {
string str = "aabcbcdbca";
cout << findSubString(str);
return 0;
}
import java.util.Arrays;
class GfG {
public static int findSubString(String str) {
int n = str.length();
int ans = n;
// to store all distinct characters
boolean[] visited = new boolean[26];
for (int i = 0; i < n; i++) {
visited[str.charAt(i) - 'a'] = true;
}
for (int i = 0; i < n; i++) {
boolean[] cur = new boolean[26];
for (int j = i; j < n; j++) {
cur[str.charAt(j) - 'a'] = true;
// if all characters are present
if (Arrays.equals(cur, visited)) {
ans = Math.min(ans, j - i + 1);
break;
}
}
}
return ans;
}
public static void main(String[] args) {
String str = "aabcbcdbca";
System.out.print(findSubString(str));
}
}
def findSubString(str):
n = len(str)
ans = n
# to store all distinct characters
visited = [False] * 26
for i in range(n):
visited[ord(str[i]) - ord('a')] = True
for i in range(n):
cur = [False] * 26
for j in range(i, n):
cur[ord(str[j]) - ord('a')] = True
# if all characters are present
if cur == visited:
ans = min(ans, j - i + 1)
break
return ans
if __name__ == "__main__":
str = "aabcbcdbca"
print(findSubString(str))
using System;
using System.Linq;
class GfG {
public static int findSubString(string str) {
int n = str.Length;
int ans = n;
// to store all distinct characters
bool[] visited = new bool[26];
for (int i = 0; i < n; i++) {
visited[str[i] - 'a'] = true;
}
for (int i = 0; i < n; i++) {
bool[] cur = new bool[26];
for (int j = i; j < n; j++) {
cur[str[j] - 'a'] = true;
// if all characters are present
if (cur.SequenceEqual(visited)) {
ans = Math.Min(ans, j - i + 1);
break;
}
}
}
return ans;
}
static void Main() {
string str = "aabcbcdbca";
Console.Write(findSubString(str));
}
}
function findSubString(str) {
let n = str.length;
let ans = n;
// to store all distinct characters
let visited = new Array(26).fill(false);
for (let i = 0; i < n; i++) {
visited[str.charCodeAt(i) - 97] = true;
}
for (let i = 0; i < n; i++) {
let cur = new Array(26).fill(false);
for (let j = i; j < n; j++) {
cur[str.charCodeAt(j) - 97] = true;
// if all characters are present
if (JSON.stringify(cur) === JSON.stringify(visited)) {
ans = Math.min(ans, j - i + 1);
break;
}
}
}
return ans;
}
let str = "aabcbcdbca";
console.log(findSubString(str));
[Expected Approach] - Using Sliding Window - O(n) Time and O(1) Space
The idea is to maintain a sliding window over the string using two pointers, expanding the window until it contains all distinct characters, then contracting it from the left to remove any redundant characters while still covering all distinct characters. By continuously updating the minimum window size seen whenever the window is valid, we achieve a linear-time solution.
Follow the below given steps:
- Use an array visited
[26] to mark all distinct characters in the input string, tracking the total in distinct. - Initialize a second array
cur[26] to count character frequencies within the current window, and a counter cnt to track how many distinct characters the window currently contains. - Set
start = 0 and iterate i from 0 to n–1:- Increment
cur[str[i] - 'a'], if this becomes 1, increment cnt. - While
cnt == distinct, update ans = min(ans, i – start + 1), then decrement cur[str[start] - 'a'], if it becomes 0, decrement cnt, and increment start to shrink the window.
- Return
ans as the length of the smallest substring containing all distinct characters.
C++
#include <bits/stdc++.h>
using namespace std;
int findSubString(string str) {
int n = str.size();
// to store all distinct characters
vector<bool> visited(26, false);
int distinct = 0;
for (int i = 0; i < n; i++) {
if (visited[str[i] - 'a'] == false) {
visited[str[i] - 'a'] = true;
distinct++;
}
}
// to store the visited of characters
// in the current window
vector<int> cur(26, 0);
int cnt = 0;
int ans = n;
int start = 0;
for(int i = 0; i < n; i++) {
// Count characters in the current
// window
cur[str[i] - 'a']++;
if(cur[str[i] - 'a'] == 1) {
cnt++;
}
// If the count becomes same as overall
while(cnt == distinct) {
ans = min(ans, i - start + 1);
// Remove characters from left
cur[str[start] - 'a']--;
if(cur[str[start] - 'a'] == 0) {
cnt--;
}
start++;
}
}
return ans;
}
int main() {
string str = "aabcbcdbca";
cout << findSubString(str);
return 0;
}
import java.util.Arrays;
class GfG {
public static int findSubString(String str) {
int n = str.length();
// to store all distinct characters
boolean[] visited = new boolean[26];
int distinct = 0;
for (int i = 0; i < n; i++) {
if (visited[str.charAt(i) - 'a'] == false) {
visited[str.charAt(i) - 'a'] = true;
distinct++;
}
}
// to store the visited of characters
// in the current window
int[] cur = new int[26];
int cnt = 0;
int ans = n;
int start = 0;
for (int i = 0; i < n; i++) {
cur[str.charAt(i) - 'a']++;
if (cur[str.charAt(i) - 'a'] == 1) {
cnt++;
}
while (cnt == distinct) {
ans = Math.min(ans, i - start + 1);
cur[str.charAt(start) - 'a']--;
if (cur[str.charAt(start) - 'a'] == 0) {
cnt--;
}
start++;
}
}
return ans;
}
public static void main(String[] args) {
String str = "aabcbcdbca";
System.out.print(findSubString(str));
}
}
def findSubString(str):
n = len(str)
# to store all distinct characters
visited = [False] * 26
distinct = 0
for i in range(n):
if visited[ord(str[i]) - ord('a')] == False:
visited[ord(str[i]) - ord('a')] = True
distinct += 1
# to store the visited of characters
# in the current window
cur = [0] * 26
cnt = 0
ans = n
start = 0
for i in range(n):
cur[ord(str[i]) - ord('a')] += 1
if cur[ord(str[i]) - ord('a')] == 1:
cnt += 1
while cnt == distinct:
ans = min(ans, i - start + 1)
cur[ord(str[start]) - ord('a')] -= 1
if cur[ord(str[start]) - ord('a')] == 0:
cnt -= 1
start += 1
return ans
if __name__ == "__main__":
str = "aabcbcdbca"
print(findSubString(str))
using System;
class GfG {
public static int findSubString(string str) {
int n = str.Length;
// to store all distinct characters
bool[] visited = new bool[26];
int distinct = 0;
for (int i = 0; i < n; i++) {
if (visited[str[i] - 'a'] == false) {
visited[str[i] - 'a'] = true;
distinct++;
}
}
// to store the visited of characters
// in the current window
int[] cur = new int[26];
int cnt = 0;
int ans = n;
int start = 0;
for (int i = 0; i < n; i++) {
cur[str[i] - 'a']++;
if (cur[str[i] - 'a'] == 1) {
cnt++;
}
while (cnt == distinct) {
ans = Math.Min(ans, i - start + 1);
cur[str[start] - 'a']--;
if (cur[str[start] - 'a'] == 0) {
cnt--;
}
start++;
}
}
return ans;
}
static void Main() {
string str = "aabcbcdbca";
Console.Write(findSubString(str));
}
}
function findSubString(str) {
let n = str.length;
// to store all distinct characters
let visited = new Array(26).fill(false);
let distinct = 0;
for (let i = 0; i < n; i++) {
if (visited[str.charCodeAt(i) - 97] == false) {
visited[str.charCodeAt(i) - 97] = true;
distinct++;
}
}
// to store the visited of characters
// in the current window
let cur = new Array(26).fill(0);
let cnt = 0;
let ans = n;
let start = 0;
for (let i = 0; i < n; i++) {
cur[str.charCodeAt(i) - 97]++;
if (cur[str.charCodeAt(i) - 97] == 1) {
cnt++;
}
while (cnt == distinct) {
ans = Math.min(ans, i - start + 1);
cur[str.charCodeAt(start) - 97]--;
if (cur[str.charCodeAt(start) - 97] == 0) {
cnt--;
}
start++;
}
}
return ans;
}
// driver code
let str = "aabcbcdbca";
console.log(findSubString(str));
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