Count Substrings with equal number of 0s, 1s and 2s

Given a string that consists of only 0s, 1s and 2s, count the number of substrings that have an equal number of 0s, 1s, and 2s.

Examples:

Input: str = “0102010”
Output:  2
Explanation: Substring str[2, 4] = “102” and substring str[4, 6] = “201” has equal number of 0, 1 and 2

Input: str = “102100211”
Output: 5

Brute Force: To solve the problem using this approach follow the below idea:

Iterate through all substrings of str using nested loops and check whether they contain equal 0,1 and 2 or not.

C++

// C++ program to find substring with equal
// number of 0's, 1's and 2's
#include <bits/stdc++.h>
using namespace std;

// Method to count number of substring which
// has equal 0, 1 and 2
long long getSubstringWithEqual012(string s) {

        vector<string> arr;
        int n = s.size();
          //generating subarrays
        for(int i=0;i<n;i++)
        {
            for(int j=i;j<n;j++)
            {
                string s1 = "";
                for(int k=i;k<=j;k++)
                {
                    s1+=s[k];
                }
                arr.push_back(s1);
            }
        }
        int count = 0;
        int countZero,countOnes,countTwo;
          //    iterating over array of all substrings
        for(int i=0;i<arr.size();i++)
        {
            countZero=0;
            countOnes=0;
            countTwo=0;
            string curs = arr[i];
            for(int j=0;j<curs.size();j++)
            {
                if(curs[j] == '0')
                countZero++;
                if(curs[j] == '1')
                countOnes++;
                if(curs[j] == '2')
                countTwo++;
            }
            // if number of ones,two and zero are equal in a substring
            if(countZero == countOnes and countOnes == countTwo)
            {
                count++;
            }
        }


        return count;
    }
//  Driver's code
int main()
{
    string str = "0102010";

    // Function call
    cout << getSubstringWithEqual012(str) << endl;
    return 0;
}

// This code is contributed by Arpit Jain

Java

// Java program to find substring with equal
// number of 0's, 1's and 2's

import java.util.*;

public class GFG {

    // Method to count number of substring which
    // has equal 0, 1 and 2
    static long getSubstringWithEqual012(String s)
    {

        ArrayList<String> arr = new ArrayList<>();
        int n = s.length();
        // generating subarrays
        for (int i = 0; i < n; i++) {
            for (int j = i; j < n; j++) {
                String s1 = "";
                for (int k = i; k <= j; k++) {
                    s1 += s.charAt(k);
                }
                arr.add(s1);
            }
        }
        int count = 0;
        int countZero, countOnes, countTwo;
        //    iterating over array of all substrings
        for (int i = 0; i < arr.size(); i++) {
            countZero = 0;
            countOnes = 0;
            countTwo = 0;
            String curs = arr.get(i);
            for (int j = 0; j < curs.length(); j++) {
                if (curs.charAt(j) == '0')
                    countZero++;
                if (curs.charAt(j) == '1')
                    countOnes++;
                if (curs.charAt(j) == '2')
                    countTwo++;
            }
            // if number of ones,two and zero are equal in a
            // substring
            if (countZero == countOnes
                && countOnes == countTwo) {
                count++;
            }
        }

        return count;
    }
    //  Driver's code
    public static void main(String[] args)
    {
        String str = "0102010";
        System.out.println(getSubstringWithEqual012(str));
    }
}

// This code is contributed by Karandeep1234

Python3

# Python3 program to find subString with equal
# number of 0's, 1's and 2's

# Method to count number of subString which
# has equal 0, 1 and 2
def getSubStringWithEqual012(s) :

    arr = [];
    n = len(s);

    # generating subarrays
    for i in range(n):
        for j in range(i, n):

            s1 = ""
            for k in range(i, 1 + j):
                s1+=s[k];

            arr.append(s1);

    count = 0;

    #    iterating over array of all subStrings
    for i in range(len(arr)):

        countZero=0;
        countOnes=0;
        countTwo=0;
        curs = arr[i];

        for j in range(len(curs)):

            if(curs[j] == '0'):
                countZero+=1;
            if(curs[j] == '1'):
                countOnes+=1;
            if(curs[j] == '2'):
                countTwo+=1;

        # if number of ones,two and zero are equal in a subString
        if(countZero == countOnes and countOnes == countTwo):
            count += 1;

    return count;

#  Driver's code
Str = "0102010";

# Function call
print(getSubStringWithEqual012(Str));

# This code is contributed by phasing17

C#

// C# program to find substring with equal
// number of 0's, 1's and 2's
using System;
using System.Collections;
using System.Collections.Generic;

public class GFG {

  // Method to count number of substring which
  // has equal 0, 1 and 2
  static long getSubstringWithEqual012(string s)
  {

    List<string> arr = new List<string>();
    int n = s.Length;
    // generating subarrays
    for (int i = 0; i < n; i++) {
      for (int j = i; j < n; j++) {
        String s1 = "";
        for (int k = i; k <= j; k++) {
          s1 += s[k];
        }
        arr.Add(s1);
      }
    }
    int count = 0;
    int countZero, countOnes, countTwo;
    //    iterating over array of all substrings
    for (int i = 0; i < arr.Count; i++) {
      countZero = 0;
      countOnes = 0;
      countTwo = 0;
      String curs = arr[i];
      for (int j = 0; j < curs.Length; j++) {
        if (curs[j] == '0')
          countZero++;
        if (curs[j] == '1')
          countOnes++;
        if (curs[j] == '2')
          countTwo++;
      }
      // if number of ones,two and zero are equal in a
      // substring
      if (countZero == countOnes
          && countOnes == countTwo) {
        count++;
      }
    }

    return count;
  }

  //  Driver's code
  public static void Main(string[] args)
  {
    string str = "0102010";
    Console.WriteLine(getSubstringWithEqual012(str));
  }
}

// This code is contributed by Karandeep1234

Javascript

// JS program to find substring with equal
// number of 0's, 1's and 2's


// Method to count number of substring which
// has equal 0, 1 and 2
function getSubstringWithEqual012(s) {

        let arr = [];
        let n = s.length;

        //generating subarrays
        for(var i=0;i<n;i++)
        {
            for(var j=i;j<n;j++)
            {
                var s1 = "";
                for(var k=i;k<=j;k++)
                {
                    s1+=s[k];
                }
                arr.push(s1);
            }
        }
        var count = 0;
        var countZero,countOnes,countTwo;
          //    iterating over array of all substrings
        for(var i=0;i<arr.length;i++)
        {
            countZero=0;
            countOnes=0;
            countTwo=0;
            var curs = arr[i];
            for(var j=0;j<curs.length;j++)
            {
                if(curs[j] == '0')
                countZero++;
                if(curs[j] == '1')
                countOnes++;
                if(curs[j] == '2')
                countTwo++;
            }
            // if number of ones,two and zero are equal in a substring
            if(countZero == countOnes && countOnes == countTwo)
            {
                count++;
            }
        }


        return count;
    }
//  Driver's code
let str = "0102010";

// Function call
console.log(getSubstringWithEqual012(str));


// This code is contributed by phasing17

Output

2

Time Complexity: O(N³)
Auxiliary Space: O(1)

Count Substrings with equal number of 0s, 1s and 2s using Hashing:

Traverse through the string and keep track of counts of 0, 1, and 2 and make a difference pair of (zeroes – ones, zeroes – twos) and increase the answer count if this difference pair is seen before and at every index increase the count of this difference pair in the map

 Follow the given steps to solve the problem:

• Declare a map to store the difference pair and three variables to store the count of 0’s, 1’s and 2’s 
• Traverse the string and keep track of the count of 0’s, 1’s, and 2’s
• At each index make a difference pair of (zeroes – ones, zeroes – twos)
• Using the map check if this pair is seen before, if it is so then increase the result count
• Then, increase the count of this pair in the map
• Return the result

Below is the implementation of the above approach:

C++

// C++ program to find substring with equal
// number of 0's, 1's and 2's
#include <bits/stdc++.h>
using namespace std;

// Method to count number of substring which
// has equal 0, 1 and 2
int getSubstringWithEqual012(string str)
{
    int N = str.length();

    // map to store, how many times a difference
    // pair has occurred previously
    map<pair<int, int>, int> mp;
    mp[make_pair(0, 0)] = 1;

    //  zc (Count of zeroes), oc(Count of 1s)
    //  and tc(count of twos)
    //  In starting all counts are zero
    int zc = 0, oc = 0, tc = 0;

    //  looping into string
    int res = 0; // Initialize result
    for (int i = 0; i < N; ++i) {

        // increasing the count of current character
        if (str[i] == '0')
            zc++;
        else if (str[i] == '1')
            oc++;
        else
            tc++; // Assuming that string doesn't contain
                  // other characters

        // making pair of differences (z[i] - o[i],
        // z[i] - t[i])
        pair<int, int> tmp = make_pair(zc - oc, zc - tc);

        // Count of previous occurrences of above pair
        // indicates that the subarrays forming from
        // every previous occurrence to this occurrence
        // is a subarray with equal number of 0's, 1's
        // and 2's
        res = res + mp[tmp];

        // Increasing the count of current difference
        // pair by 1
        mp[tmp]++;
    }

    return res;
}

//  Driver's code
int main()
{
    string str = "0102010";

    // Function call
    cout << getSubstringWithEqual012(str) << endl;
    return 0;
}

Java

// Java program to find substring with equal
// number of 0's, 1's and 2's

import java.io.*;
import java.util.*;

class GFG {

    // Method to count number of substring which
    // has equal 0, 1 and 2
    private static int getSubstringWithEqual012(String str)
    {

        // map to store, how many times a difference
        // pair has occurred previously (key = diff1 *
        // diff2)
        HashMap<String, Integer> map = new HashMap<>();
        map.put("0*0", 1);

        //  zc (Count of zeroes), oc(Count of 1s)
        //  and tc(count of twos)
        //  In starting all counts are zero
        int zc = 0, oc = 0, tc = 0;

        int ans = 0;

        //  looping into string
        for (int i = 0; i < str.length(); i++) {

            // Increasing the count of current character
            if (str.charAt(i) == '0')
                zc++;
            else if (str.charAt(i) == '1')
                oc++;
            else
                tc++;

            // making key of differences (z[i] - o[i],
            // z[i] - t[i])
            String key = (zc - oc) + "*" + (zc - tc);

            // Count of previous occurrences of above pair
            // indicates that the subarrays forming from
            // every previous occurrence to this occurrence
            // is a subarray with equal number of 0's, 1's
            // and 2's
            ans += map.getOrDefault(key, 0);

            map.put(key, map.getOrDefault(key, 0) + 1);
        }

        // increasing the count of current difference
        // pair by 1
        return ans;
    }

    // Driver's Code
    public static void main(String[] args)
    {
        String str = "0102010";

          // Function call
        System.out.println(getSubstringWithEqual012(str));
    }
}

Python3

# Python3 program to find substring with equal
# number of 0's, 1's and 2's

# Method to count number of substring which
# has equal 0, 1 and 2
def getSubstringWithEqual012(string):
    N = len(string)

    # map to store, how many times a difference
    # pair has occurred previously
    mp = dict()
    mp[(0, 0)] = 1

    # zc (Count of zeroes), oc(Count of 1s)
    # and tc(count of twos)
    # In starting all counts are zero
    zc, oc, tc = 0, 0, 0

    # looping into string
    res = 0 # Initialize result
    for i in range(N):

        # increasing the count of current character
        if string[i] == '0':
            zc += 1
        elif string[i] == '1':
            oc += 1
        else:
            tc += 1 # Assuming that string doesn't contain
                    # other characters

        # making pair of differences (z[i] - o[i],
        # z[i] - t[i])
        tmp = (zc - oc, zc - tc)

        # Count of previous occurrences of above pair
        # indicates that the subarrays forming from
        # every previous occurrence to this occurrence
        # is a subarray with equal number of 0's, 1's
        # and 2's
        if tmp not in mp:
            res += 0
        else:
            res += mp[tmp]

        # increasing the count of current difference
        # pair by 1
        if tmp in mp:
            mp[tmp] += 1
        else:
            mp[tmp] = 1

    return res

# Driver's Code
if __name__ == "__main__":
    string = "0102010"
    print(getSubstringWithEqual012(string))

# This code is contributed by
# sanjeev2552

C#

// C# program to implement the approach

using System;
using System.Collections.Generic;

class GFG
{
    // Method to count number of substring which
    // has equal 0, 1 and 2
    private static int getSubstringWithEqual012(string str)
    {
        // dictionary to store, how many times a difference
        // pair has occurred previously (key = diff1 *
        // diff2)
        Dictionary<string, int> map = new Dictionary<string, int>();
        map.Add("0*0", 1);

        //  zc (Count of zeroes), oc(Count of 1s)
        //  and tc(count of twos)
        //  In starting all counts are zero
        int zc = 0, oc = 0, tc = 0;

        int ans = 0;

        //  looping into string
        for (int i = 0; i < str.Length; i++)
        {
            // Increasing the count of current character
            if (str[i] == '0')
                zc++;
            else if (str[i] == '1')
                oc++;
            else
                tc++;

            // making key of differences (z[i] - o[i],
            // z[i] - t[i])
            string key = (zc - oc) + "*" + (zc - tc);

            // Count of previous occurrences of above pair
            // indicates that the subarrays forming from
            // every previous occurrence to this occurrence
            // is a subarray with equal number of 0's, 1's
            // and 2's
            if (map.ContainsKey(key))
                ans += map[key];
            else
                map.Add(key, 1);
        }

        // returning the count of subarray with equal
        // number of 0's, 1's and 2's
        return ans;
    }

    // Driver's Code
    public static void Main(string[] args)
    {
        string str = "0102010";

        // Function call
        Console.WriteLine(getSubstringWithEqual012(str));
    }
}

Javascript

function getSubstringWithEqual012(string) {
    const N = string.length;
    // map to store, how many times a difference pair has occurred previously
    const mp = {};
    mp[[0, 0]] = 1;

    // zc (Count of zeroes), oc(Count of 1s) and tc(count of twos)
    // In starting all counts are zero
    let zc = 0, oc = 0, tc = 0;

    // looping into string
    let res = 0; // Initialize result
    for (let i = 0; i < N; i++) {
        // increasing the count of current character
        if (string[i] === '0') {
            zc += 1;
        } else if (string[i] === '1') {
            oc += 1;
        } else {
            tc += 1; // Assuming that string doesn't contain other characters
        }

        // making pair of differences (z[i] - o[i], z[i] - t[i])
        const tmp = [zc - oc, zc - tc];

        // Count of previous occurrences of above pair
        // indicates that the subarrays forming from
        // every previous occurrence to this occurrence is a
        // subarray with equal number of 0's, 1's
        // and 2's
        if (!mp.hasOwnProperty(tmp)) {
            res += 0;
        } else {
            res += mp[tmp];
        }

        // increasing the count of current difference pair by 1
        if (mp.hasOwnProperty(tmp)) {
            mp[tmp] += 1;
        } else {
            mp[tmp] = 1;
        }
    }
    return res;
}

// Driver's Code
console.log(getSubstringWithEqual012("0102010"));

Output

2

Time Complexity: O(N * log N)
Auxiliary Space: O(N)