Smallest window containing 0, 1 and 2
Last Updated :
30 Mar, 2023
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++
#include <bits/stdc++.h>
using namespace std;
int smallestSubstring(string S)
{
int res = INT_MAX;
bool zero = false, one = false, two = false;
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;
}
if (zero and one and two)
res = min(res,
max({ zeroindex, oneindex, twoindex })
- min({ zeroindex, oneindex, twoindex }));
}
if (res == INT_MAX)
return -1;
return res + 1;
}
int main()
{
string S = "01212";
cout << smallestSubstring(S);
return 0;
}
|
C
#include <limits.h> // to use INT_MAX
#include <stdbool.h> //to use bool variable
#include <stdio.h>
int min_two(int a, int b)
{
if (a <= b)
return a;
else
return b;
}
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;
}
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;
}
int smallestSubstring(char S[], int n)
{
int res = INT_MAX;
bool zero = 0, one = 0, two = 0;
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;
}
if (zero && one && two)
res = min_two( res,
max_three(zeroindex, oneindex, twoindex)
- min_three(zeroindex, oneindex, twoindex));
}
if (res == INT_MAX)
return -1;
return res + 1;
}
int main()
{
int n = 5;
char S[] = "01212";
int ans = smallestSubstring(S, n);
printf("%d", ans);
return 0;
}
|
Java
import java.util.*;
class GFG {
public static int smallestSubstring(String S)
{
int res = Integer.MAX_VALUE;
boolean zero = false, one = false, two = false;
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;
}
if (zero && one && two)
res = Math.min( res,
Math.max(zeroindex,Math.max(oneindex, twoindex))
- Math.min( zeroindex,Math.min(oneindex, twoindex)));
}
if (res == Integer.MAX_VALUE)
return -1;
return res + 1;
}
public static void main(String[] args)
{
String S = "01212";
System.out.print(smallestSubstring(S));
}
}
|
Python3
def smallestSubstring(S):
res = 999999
zero = 0
one = 0
two = 0
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
if (zero and one and two):
res = min(res,
max([zeroindex, oneindex, twoindex])
- min([zeroindex, oneindex, twoindex]))
if (res == 999999):
return -1
return res + 1
S = "01212"
print(smallestSubstring(S))
|
C#
using System;
public class GFG {
static int smallestSubstring(string S)
{
int res = Int32.MaxValue;
bool zero = false, one = false, two = false;
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;
}
if (zero && one && two)
res = Math.Min(res,
Math.Max( zeroindex, Math.Max(oneindex, twoindex))
- Math.Min( zeroindex, Math.Min(oneindex, twoindex)));
}
if (res == Int32.MaxValue)
return -1;
return res + 1;
}
static public void Main()
{
string S = "01212";
Console.Write(smallestSubstring(S));
}
}
|
Javascript
<script>
const INT_MAX = 2147483647;
const smallestSubstring = (S) => {
let res = INT_MAX;
let zero = false, one = false, two = false;
let zeroindex, oneindex, twoindex;
for (let 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;
}
if (zero && one && two)
res = Math.min(res,
Math.max(...[zeroindex, oneindex, twoindex])
- Math.min(...[zeroindex, oneindex, twoindex]));
}
if (res == INT_MAX)
return -1;
return res + 1;
}
let S = "01212";
document.write(smallestSubstring(S));
</script>
|
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
- Initialize three pointers i, j, and k to 0.
- Initialize a variable min_len to the maximum integer value.
- 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
- 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.
- 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 <bits/stdc++.h>
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[3] = {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));
}
}
|
Python3
def smallestSubstring(S):
n, i, j, k, cnt, min_len = len(S), 0, 0, 0, 0, float("inf")
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 -1 if min_len == float("inf") else min_len
S = "01212"
print(smallestSubstring(S))
|
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[3];
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) {
let n = S.length,
i = 0,
j = 0,
k = 0,
cnt = 0,
min_len = Infinity;
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 min_len == Infinity ? -1 : min_len;
}
let S = "01212";
console.log(smallestSubstring(S));
|
Time Complexity: O(N)
Auxiliary Space: O(1)
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