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

05 Jul, 2022
Given a string which consists of only 0s, 1s or 2s, count the number of substrings that have 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
A simple solution is to iterate through all substring of str and checking whether they contain equal 0,1 and 2 or not. Total number of substring of str is O(n2) checking each substring for count takes O(n) times, So total time taken to solve this problem is O(n3) time with brute-force approach.
An efficient solution is to keep track of counts of 0, 1, and 2.  
Let zc[i] denotes number of zeros between index 1 and i
    oc[i] denotes number of ones between index 1 and i
    tc[i] denotes number of twos between index 1 and i
for substring str[i, j] to be counted in result we should have :
    zc[i] – zc[j-1] = oc[i] – oc[j-1] = tc[i] - tc[j-1]
we can write above relation as follows :
z[i] – o[i] = z[j-1] – o[j-1]    and
z[i] – t[i] = z[j-1] – t[j-1]
The above relations can be tracked while looping in string, at each index i we’ll calculate this difference pair and we’ll check how many time this difference pair has previously occurred and we’ll add that count to our result, for keeping track of previous difference pair we have used map in below code. Total time complexity of this solution is O(n log n) considering the fact that map operations, like search and insert take O(Log n) time. 
Implementation:
C++
Java
Python3
// 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 2int 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 code to test above methodint main(){    string str = "0102010";    cout << getSubstringWithEqual012(str) << endl;    return 0;}
// Java program to find substring with equal// number of 0's, 1's and 2'simport 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 Code    public static void main(String[] args)    {        // Input        String str = "0102010";        System.out.println(getSubstringWithEqual012(str));    }}
# 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 2def 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        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])        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 Codeif __name__ == "__main__":    string = "0102010"    print(getSubstringWithEqual012(string)) # This code is contributed by# sanjeev2552
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