# 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:2Explanation: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` |

**Time Complexity: **O(N^{3})** ****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` |

**Output**

2

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

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