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# Smallest window containing 0, 1 and 2

Given a string S of size N consisting of the characters 0, 1 and 2, the task is to find the length of the smallest substring of string S that contains all the three characters 0, 1 and 2. If no such substring exists, then return -1.

Examples:

Input: S = “01212”
Output: 3
Explanation: The substring 012 is the smallest substring
that contains the characters 0, 1 and 2.

Input:  S = “12121”
Output: -1
Explanation:  As the character 0 is not present in the
string S, therefore no substring containing
all the three characters 0, 1 and 2
exists. Hence, the answer is -1 in this case.

Approach: The idea of the approach is as mentioned below:

Use three pointers to store the indices of the elements 0, 1 and 2. When all the three elements are found, the difference distance between the maximum of them and the minimum of them is the minimum size window.
Keep updating the pointers whenever any of them is found again and calculate the size of the new window.

Follow the illustration below for a better understanding.

Illustration:

Consider S = “01212” and the three pointers to be zeroindex, oneindex and twoindex and set all of them to -1 initially.

When i = 0:
=> S[i] = ‘0’. zeroindex = 0, oneindex = -1, twoindex = -1
=> All of the values are not found. So no window is possible

When i = 1:
=> S[i] = ‘1’. zeroindex = 0, oneindex = 1, twoindex = -1
=> All of the values are not found. So no window is possible

When i = 2:
=> S[i] = ‘2’. zeroindex = 0, oneindex = 1, twoindex = 2
=> All of the values are found.
=> Maximum is twoindex = 2. Minimum is zeroindex = 0.
=> So window size = (2 – 0 + 1) = 3.
=> Minimum window size = 3

When i = 3:
=> S[i] = ‘1’. zeroindex = 0, oneindex = 3, twoindex = 2
=> All of the values are found.
=> Maximum is oneindex = 3. Minimum is zeroindex = 0.
=> So window size = (3 – 0 + 1) = 4.
=> Minimum window size = min (3, 4) = 3

When i = 4:
=> S[i] = ‘2’. zeroindex = 0, oneindex = 3, twoindex = 4
=> All of the values are found.
=> Maximum is twoindex = 4. Minimum is zeroindex = 0.
=> So window size = (4 – 0 + 1) = 5.
=> Minimum window size = min(3, 5) = 3

So the size of the smallest window is 3

Follow the below steps to solve the problem:

• Take three variable zero, one and two to check if 0, 1 and 2 are found in the window or not.
• Take three variables zeroindex, oneindex and twoindex which will store indexes of 0, 1 and 2 when we encounter them.
• Run the for loop for the whole length of String:
• Update the indices of the values encountered.
• Update the length of the window, if three of them are found.
• Length will be the difference between the maximum and the minimum of the indexes of 0, 1 and 2.
• And if all three values i.e., 0, 1, 2 are not found after the traversal is over then in that case return ‘-1’.

Below is the implementation of the above approach:

## C++

 `// C++ program for above approach`   `#include ` `using` `namespace` `std;`   `// Function to find the length of the smallest substring` `int` `smallestSubstring(string S)` `{` `    ``int` `res = INT_MAX;` `    ``// To check 0, 1 and 2` `    ``bool` `zero = ``false``, one = ``false``, two = ``false``;` `    ``// To store indexes of 0, 1 and 2` `    ``int` `zeroindex, oneindex, twoindex;` `    ``for` `(``int` `i = 0; i < S.length(); i++) {` `        ``if` `(S[i] == ``'0'``) {` `            ``zero = ``true``;` `            ``zeroindex = i;` `        ``}` `        ``else` `if` `(S[i] == ``'1'``) {` `            ``one = ``true``;` `            ``oneindex = i;` `        ``}` `        ``else` `if` `(S[i] == ``'2'``) {` `            ``two = ``true``;` `            ``twoindex = i;` `        ``}`   `        ``// Calculating length` `        ``if` `(zero and one and two)` `            ``res = min(res,` `                      ``max({ zeroindex, oneindex, twoindex })` `                          ``- min({ zeroindex, oneindex, twoindex }));` `    ``}`   `    ``// In case if there is no substring that contains 0,1 and 2` `    ``if` `(res == INT_MAX)` `        ``return` `-1;` `    ``return` `res + 1;` `}`   `// Driver Code` `int` `main()` `{` `    ``string S = ``"01212"``;`   `    ``// Function call` `    ``cout << smallestSubstring(S);` `    ``return` `0;` `}`   `// This code is contributed by Aditya Kumar (adityakumar129)`

## C

 `// C program for above approach`   `#include // to use INT_MAX` `#include //to use bool variable` `#include `   `// function to calculate max of two numbers` `int` `min_two(``int` `a, ``int` `b)` `{` `    ``if` `(a <= b)` `        ``return` `a;` `    ``else` `        ``return` `b;` `}`   `// function to calculate max of three numbers` `int` `max_three(``int` `a, ``int` `b, ``int` `c)` `{` `    ``if` `(a >= b && a >= c)` `        ``return` `a;` `    ``else` `if` `(b > a && b >= c)` `        ``return` `b;` `    ``else` `if` `(c > a && c > b)` `        ``return` `c;` `}`   `// function to calculate min of three numbers` `int` `min_three(``int` `a, ``int` `b, ``int` `c)` `{` `    ``if` `(a <= b && a <= c)` `        ``return` `a;` `    ``else` `if` `(b < a && b <= c)` `        ``return` `b;` `    ``else` `if` `(c < a && c < b)` `        ``return` `c;` `}`   `// Function to find the length of` `// the smallest substring` `int` `smallestSubstring(``char` `S[], ``int` `n)` `{` `    ``int` `res = INT_MAX;`   `    ``// To check 0, 1 and 2` `    ``bool` `zero = 0, one = 0, two = 0;`   `    ``// To store indexes of 0, 1 and 2` `    ``int` `zeroindex, oneindex, twoindex;` `    ``for` `(``int` `i = 0; i < n; i++) {` `        ``if` `(S[i] == ``'0'``) {` `            ``zero = ``true``;` `            ``zeroindex = i;` `        ``}` `        ``else` `if` `(S[i] == ``'1'``) {` `            ``one = ``true``;` `            ``oneindex = i;` `        ``}` `        ``else` `if` `(S[i] == ``'2'``) {` `            ``two = ``true``;` `            ``twoindex = i;` `        ``}`   `        ``// Calculating length` `        ``if` `(zero && one && two)` `            ``res = min_two( res,` `                     ``max_three(zeroindex, oneindex, twoindex)` `                    ``- min_three(zeroindex, oneindex, twoindex));` `    ``}`   `    ``// In case if there is no substring that contains 0,1 and 2` `    ``if` `(res == INT_MAX)` `        ``return` `-1;` `    ``return` `res + 1;` `}`   `// Driver Code` `int` `main()` `{` `    ``int` `n = 5; ``// size of string` `    ``char` `S[] = ``"01212"``;`   `    ``int` `ans = smallestSubstring(S, n);` `    ``// Function call` `    ``printf``(``"%d"``, ans);` `    ``return` `0;` `}`   `// This code is contributed by Aditya Kumar (adityakumar129)`

## Java

 `// Java program for above approach` `import` `java.util.*;` `class` `GFG {`   `    ``// Function to find the length of the smallest substring` `    ``public` `static` `int` `smallestSubstring(String S)` `    ``{` `        ``int` `res = Integer.MAX_VALUE;`   `        ``// To check 0, 1 and 2` `        ``boolean` `zero = ``false``, one = ``false``, two = ``false``;`   `        ``// To store indexes of 0, 1 and 2` `        ``int` `zeroindex = ``0``, oneindex = ``0``, twoindex = ``0``;` `        ``for` `(``int` `i = ``0``; i < S.length(); i++) {` `            ``if` `(S.charAt(i) == ``'0'``) {` `                ``zero = ``true``;` `                ``zeroindex = i;` `            ``}` `            ``else` `if` `(S.charAt(i) == ``'1'``) {` `                ``one = ``true``;` `                ``oneindex = i;` `            ``}` `            ``else` `if` `(S.charAt(i) == ``'2'``) {` `                ``two = ``true``;` `                ``twoindex = i;` `            ``}`   `            ``// Calculating length` `            ``if` `(zero && one && two)` `                ``res = Math.min( res,` `                    ``Math.max(zeroindex,Math.max(oneindex, twoindex))` `                  ``- Math.min( zeroindex,Math.min(oneindex, twoindex)));` `        ``}`   `        ``// In case if there is no substring that contains 0,1 and 2` `        ``if` `(res == Integer.MAX_VALUE)` `            ``return` `-``1``;` `        ``return` `res + ``1``;` `    ``}`   `    ``// Driver Code` `    ``public` `static` `void` `main(String[] args)` `    ``{` `        ``String S = ``"01212"``;` `        ``// Function call` `        ``System.out.print(smallestSubstring(S));` `    ``}` `}`   `// This code is contributed by Aditya Kumar (adityakumar129)`

## Python3

 `# Python code for the above approach`   `# Function to find the length of the smallest substring`     `def` `smallestSubstring(S):` `    ``res ``=` `999999`   `    ``# To check 0, 1 and 2` `    ``zero ``=` `0` `    ``one ``=` `0` `    ``two ``=` `0`   `    ``# To store indexes of 0, 1 and 2` `    ``zeroindex ``=` `0` `    ``oneindex ``=` `0` `    ``twoindex ``=` `0` `    ``for` `i ``in` `range``(``len``(S)):` `        ``if` `(S[i] ``=``=` `'0'``):` `            ``zero ``=` `1` `            ``zeroindex ``=` `i`   `        ``elif` `(S[i] ``=``=` `'1'``):` `            ``one ``=` `1` `            ``oneindex ``=` `i`   `        ``elif` `(S[i] ``=``=` `'2'``):` `            ``two ``=` `1` `            ``twoindex ``=` `i`   `        ``# Calculating length` `        ``if` `(zero ``and` `one ``and` `two):` `            ``res ``=` `min``(res,` `                      ``max``([zeroindex, oneindex, twoindex])` `                      ``-` `min``([zeroindex, oneindex, twoindex]))`   `    ``# In case if there is no substring that contains 0,1 and 2` `    ``if` `(res ``=``=` `999999``):` `        ``return` `-``1` `    ``return` `res ``+` `1`     `# Driver Code` `S ``=` `"01212"`   `# Function call` `print``(smallestSubstring(S))`   `# This code is contributed by Aditya Kumar (adityakumar129)`

## C#

 `// C# program for above approach` `using` `System;`   `public` `class` `GFG {`   `    ``// Function to find the length of the smallest substring` `    ``static` `int` `smallestSubstring(``string` `S)` `    ``{` `        ``int` `res = Int32.MaxValue;`   `        ``// To check 0, 1 and 2` `        ``bool` `zero = ``false``, one = ``false``, two = ``false``;`   `        ``// To store indexes of 0, 1 and 2` `        ``int` `zeroindex = 0, oneindex = 0, twoindex = 0;` `        ``for` `(``int` `i = 0; i < S.Length; i++) {` `            ``if` `(S[i] == ``'0'``) {` `                ``zero = ``true``;` `                ``zeroindex = i;` `            ``}` `            ``else` `if` `(S[i] == ``'1'``) {` `                ``one = ``true``;` `                ``oneindex = i;` `            ``}` `            ``else` `if` `(S[i] == ``'2'``) {` `                ``two = ``true``;` `                ``twoindex = i;` `            ``}`   `            ``// Calculating length` `            ``if` `(zero && one && two)` `                ``res = Math.Min(res,` `                    ``Math.Max( zeroindex, Math.Max(oneindex, twoindex))` `                  ``- Math.Min( zeroindex, Math.Min(oneindex, twoindex)));` `        ``}`   `        ``// In case if there is no substring that contains 0,1 and 2` `        ``if` `(res == Int32.MaxValue)` `            ``return` `-1;` `        ``return` `res + 1;` `    ``}`   `    ``// Driver Code` `    ``static` `public` `void` `Main()` `    ``{` `        ``string` `S = ``"01212"``;`   `        ``// Function call` `        ``Console.Write(smallestSubstring(S));` `    ``}` `}`   `// This code is contributed by Aditya Kumar (adityakumar129)`

## Javascript

 ``

Output

`3`

Time Complexity: O(N), as we are using a loop to traverse N times.
Auxiliary Space: O(1), as we are not using any extra space.

Another Approach: using a sliding window

1. Initialize three pointers i, j, and k to 0.
2. Initialize a variable min_len to the maximum integer value.
3. Initialize an array freq[] to 0 of size 3 to keep track of the frequency of each character in the current window.
Traverse the string S using the pointer k:
a. Increment the frequency of the character S[k] in the freq[] array.
b. If freq[S[k]-‘0’] equals 1 (i.e., this character was not in the window before), increment a counter cnt.
c. If cnt equals 3, then we have found a window containing all three characters. Now, we can try to shrink the window from the left by incrementing i
4. until the frequency of any character becomes zero. While doing this, we also update the freq[] array and decrement the cnt if the frequency of any character becomes zero.
d. After shrinking the window, update the value of min_len if the current window size is smaller than min_len.
5. Return min_len if it is less than the maximum integer value, else return -1.

Below is the implementation of the above approach:

## C++

 `#include ` `using` `namespace` `std;`   `int` `smallestSubstring(string S) {` `    ``int` `n = S.length(), i = 0, j = 0, k = 0, cnt = 0, min_len = INT_MAX;` `    ``int` `freq = {0};` `    ``while` `(k < n) {` `        ``freq[S[k]-``'0'``]++;` `        ``if` `(freq[S[k]-``'0'``] == 1) cnt++;` `        ``if` `(cnt == 3) {` `            ``while` `(freq[S[i]-``'0'``] > 1) {` `                ``freq[S[i]-``'0'``]--;` `                ``i++;` `            ``}` `            ``min_len = min(min_len, k-i+1);` `            ``freq[S[i]-``'0'``]--;` `            ``i++;` `            ``cnt--;` `        ``}` `        ``k++;` `    ``}` `    ``return` `(min_len == INT_MAX) ? -1 : min_len;` `}`   `int` `main() {` `    ``string S = ``"01212"``;` `    ``cout << smallestSubstring(S) << endl;` `    ``return` `0;` `}`

## Java

 `import` `java.util.Arrays;`   `public` `class` `Main {`   `    ``public` `static` `int` `smallestSubstring(String S) {` `        ``int` `n = S.length(), i = ``0``, j = ``0``, k = ``0``, cnt = ``0``, min_len = Integer.MAX_VALUE;` `        ``int``[] freq = ``new` `int``[``3``];` `        ``Arrays.fill(freq, ``0``);` `        ``while` `(k < n) {` `            ``freq[S.charAt(k)-``'0'``]++;` `            ``if` `(freq[S.charAt(k)-``'0'``] == ``1``) cnt++;` `            ``if` `(cnt == ``3``) {` `                ``while` `(freq[S.charAt(i)-``'0'``] > ``1``) {` `                    ``freq[S.charAt(i)-``'0'``]--;` `                    ``i++;` `                ``}` `                ``min_len = Math.min(min_len, k-i+``1``);` `                ``freq[S.charAt(i)-``'0'``]--;` `                ``i++;` `                ``cnt--;` `            ``}` `            ``k++;` `        ``}` `        ``return` `(min_len == Integer.MAX_VALUE) ? -``1` `: min_len;` `    ``}`   `    ``public` `static` `void` `main(String[] args) {` `        ``String S = ``"01212"``;` `        ``System.out.println(smallestSubstring(S));` `    ``}` `}` `// This code is contributed by Prajwal Kandekar`

## Python3

 `# Python program to find the smallest ` `# substring containing all unique characters` `def` `smallestSubstring(S):` `  `  `    ``# Initialize variables` `    ``n, i, j, k, cnt, min_len ``=` `len``(S), ``0``, ``0``, ``0``, ``0``, ``float``(``"inf"``)` `    `  `    ``# Frequency array` `    ``freq ``=` `[``0``] ``*` `3` `    `  `    ``while` `k < n:` `        ``freq[``int``(S[k])] ``+``=` `1` `        ``if` `freq[``int``(S[k])] ``=``=` `1``:` `            ``cnt ``+``=` `1` `    `  `        ``if` `cnt ``=``=` `3``:` `            ``while` `freq[``int``(S[i])] > ``1``:` `                ``freq[``int``(S[i])] ``-``=` `1` `                ``i ``+``=` `1` `            ``min_len ``=` `min``(min_len, k``-``i``+``1``)` `            ``freq[``int``(S[i])] ``-``=` `1` `            ``i ``+``=` `1` `            ``cnt ``-``=` `1` `        ``k ``+``=` `1` `    `  `    ``# Return minimum length substring or -1 if no such substring exists` `    ``return` `-``1` `if` `min_len ``=``=` `float``(``"inf"``) ``else` `min_len`   `# Driver code` `S ``=` `"01212"`   `# Function call` `print``(smallestSubstring(S))`   `# This code is contributed by phasing17`

## C#

 `using` `System;`   `public` `class` `MainClass {` `  ``public` `static` `int` `SmallestSubstring(``string` `s) {` `    ``int` `n = s.Length, i = 0, j = 0, k = 0, cnt = 0, min_len = ``int``.MaxValue;` `    ``int``[] freq = ``new` `int``;` `    ``Array.Fill(freq, 0);`   `    ``while` `(k < n) {` `      ``freq[s[k]-``'0'``]++;` `      ``if` `(freq[s[k]-``'0'``] == 1) cnt++;` `      ``if` `(cnt == 3) {` `        ``while` `(freq[s[i]-``'0'``] > 1) {` `          ``freq[s[i]-``'0'``]--;` `          ``i++;` `        ``}` `        ``min_len = Math.Min(min_len, k-i+1);` `        ``freq[s[i]-``'0'``]--;` `        ``i++;` `        ``cnt--;` `      ``}` `      ``k++;` `    ``}` `    ``return` `(min_len == ``int``.MaxValue) ? -1 : min_len;` `  ``}`   `  ``public` `static` `void` `Main(``string``[] args) {` `    ``string` `s = ``"01212"``;` `    ``Console.WriteLine(SmallestSubstring(s));` `  ``}` `}`

## Javascript

 `function` `smallestSubstring(S) {` `  ``// Initialize variables` `  ``let n = S.length,` `    ``i = 0,` `    ``j = 0,` `    ``k = 0,` `    ``cnt = 0,` `    ``min_len = Infinity;`   `  ``// Frequency array` `  ``let freq = [0, 0, 0];`   `  ``while` `(k < n) {` `    ``freq[parseInt(S[k])] += 1;` `    ``if` `(freq[parseInt(S[k])] == 1) {` `      ``cnt += 1;` `    ``}`   `    ``if` `(cnt == 3) {` `      ``while` `(freq[parseInt(S[i])] > 1) {` `        ``freq[parseInt(S[i])] -= 1;` `        ``i += 1;` `      ``}` `      ``min_len = Math.min(min_len, k - i + 1);` `      ``freq[parseInt(S[i])] -= 1;` `      ``i += 1;` `      ``cnt -= 1;` `    ``}` `    ``k += 1;` `  ``}`   `  ``// Return minimum length substring or -1 if no such substring exists` `  ``return` `min_len == Infinity ? -1 : min_len;` `}`   `// Driver code` `let S = ``"01212"``;`   `// Function call` `console.log(smallestSubstring(S));`

Output

`3`

Time Complexity: O(N)

Auxiliary Space: O(1)

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