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(n^{2}) checking each substring for count takes O(n) times, So total time taken to solve this problem is O(n^{3}) 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.

## 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 code to test above method ` `int` `main() ` `{ ` ` ` `string str = ` `"0102010"` `; ` ` ` `cout << getSubstringWithEqual012(str) << endl; ` ` ` `return` `0; ` `} ` |

## 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 Code ` `if` `__name__ ` `=` `=` `"__main__"` `: ` ` ` `string ` `=` `"0102010"` ` ` `print` `(getSubstringWithEqual012(string)) ` ` ` `# This code is conributed by ` `# sanjeev2552 ` |

Output:

2

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