Count of substrings with equal ratios of 0s and 1s till ith index in given Binary String

Given a binary string S, the task is to print the maximum number of substrings with equal ratios of 0s and 1s till the ith index from the start.

Examples:

Input: S = “110001”
Output: {1, 2, 1, 1, 1, 2}
Explanation: The given string can be partitioned into the following equal substrings:

• Valid substrings upto index 0 : “1”, possible no. of substrings = 1
• Valid substrings upto index 1 : “1”, “B”, possible no. of substrings = 2
• Valid substrings upto index 2 : “110”, possible no. of substrings = 1
• Valid substrings upto index 3 : “1100”, possible no. of substrings = 1
• Valid substrings upto index 4 : “11000”, possible no. of substrings = 1
• Valid substrings upto index 5 : “1100”, “01”, possible no. of substrings = 2
  

Input: S = “010100001”
Output: {1, 1, 1, 2, 1, 2, 1, 1, 3}

Approach: The task can be solved using mathematical concepts & HashMap to store the frequencies of pairs Follow the below steps to solve the problem:

1. Create two prefix arrays for occurrences of 0s and 1s say pre0[] and pre1[] respectively.
2. For each prefix, label it with a pair (a, b) where a = frequency of ‘0’ and b = frequency of ‘1’ in this prefix. Divide a and b by gcd(a, b) to get the simplest form.
3. Iterate over all prefixes, and store the answer for the prefix as the current number of occurrences of this pair.

Below is the implementation of the above approach:

C++

// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;

// Function to retuan a prefix array of
// equal partitions of the given string
// consisting of 0s and 1s
void equalSubstrings(string s)
{
    // Length of the string
    int n = s.size();

    // Prefix arrays for 0s and 1s
    int pre0[n] = { 0 }, pre1[n] = { 0 };

    // If character at index 0 is 0
    if (s[0] == '0') {
        pre0[0] = 1;
    }

    // If character at index 0 is 1
    else {
        pre1[0] = 1;
    }

    // Filling the prefix arrays
    for (int i = 1; i < n; i++) {
        if (s[i] == '0') {
            pre0[i] = pre0[i - 1] + 1;
            pre1[i] = pre1[i - 1];
        }
        else {
            pre0[i] = pre0[i - 1];
            pre1[i] = pre1[i - 1] + 1;
        }
    }

    // Vector to store the answer
    vector<int> ans;

    // Map to store the ratio
    map<pair<int, int>, int> mp;

    for (int i = 0; i < n; i++) {
        // Find the gcd of pre0[i] and pre1[i]
        int x = __gcd(pre0[i], pre1[i]);

        // Converting the elements in
        // simplest form
        int l = pre0[i] / x, r = pre1[i] / x;

        // Update the value in map
        mp[{ l, r }]++;

        // Store this in ans
        ans.push_back(mp[{ l, r }]);
    }
    // Return the ans vector
    for (auto i : ans)
        cout << i << " ";
}

// Driver Code
int main()
{
    string s = "001110";
    equalSubstrings(s);

    return 0;
}

Java

// Java program for the above approach
import java.util.ArrayList;
import java.util.Arrays;
import java.util.HashMap;

class GFG
{

    // Function to retuan a prefix array of
    // equal partitions of the given string
    // consisting of 0s and 1s
    public static void equalSubstrings(String s)
    {

        // Length of the string
        int n = s.length();

        // Prefix arrays for 0s and 1s
        int[] pre0 = new int[n];
        int[] pre1 = new int[n];

        Arrays.fill(pre0, 0);
        Arrays.fill(pre1, 0);

        // If character at index 0 is 0
        if (s.charAt(0) == '0') {
            pre0[0] = 1;
        }

        // If character at index 0 is 1
        else {
            pre1[0] = 1;
        }

        // Filling the prefix arrays
        for (int i = 1; i < n; i++) {
            if (s.charAt(i) == '0') {
                pre0[i] = pre0[i - 1] + 1;
                pre1[i] = pre1[i - 1];
            } else {
                pre0[i] = pre0[i - 1];
                pre1[i] = pre1[i - 1] + 1;
            }
        }

        // Vector to store the answer
        ArrayList<Integer> ans = new ArrayList<Integer>();

        // Map to store the ratio
        HashMap<String, Integer> mp = new HashMap<String, Integer>();

        for (int i = 0; i < n; i++)
        {

            // Find the gcd of pre0[i] and pre1[i]
            int x = __gcd(pre0[i], pre1[i]);

            // Converting the elements in
            // simplest form
            int l = pre0[i] / x, r = pre1[i] / x;

            String key = l + "," + r;

            // Update the value in map
            if (mp.containsKey(key))
                mp.put(key, mp.get(key) + 1);
            else
                mp.put(key, 1);

            // Store this in ans
            ans.add(mp.get(key));
        }

        // Return the ans vector
        for (int i : ans)
            System.out.print(i + " ");
    }

    public static int __gcd(int a, int b) {
        if (b == 0)
            return a;
        return __gcd(b, a % b);
    }

    // Driver Code
    public static void main(String args[]) {
        String s = "001110";
        equalSubstrings(s);

    }
}

// This code is contributed by gfgking.

Python3

# Python program for the above approach
def __gcd(a, b):
    if (b == 0):
        return a
    return __gcd(b, a % b)

# Function to retuan a prefix array of
# equal partitions of the given string
# consisting of 0s and 1s
def equalSubstrings(s):

    # Length of the string
    n = len(s)

    # Prefix arrays for 0s and 1s
    pre0 = [0] * n
    pre1 = [0] * n

    # If character at index 0 is 0
    if (s[0] == '0'):
        pre0[0] = 1

    # If character at index 0 is 1
    else:
        pre1[0] = 1

    # Filling the prefix arrays
    for i in range(1, n):
        if (s[i] == '0'):
            pre0[i] = pre0[i - 1] + 1
            pre1[i] = pre1[i - 1]
        else:
            pre0[i] = pre0[i - 1]
            pre1[i] = pre1[i - 1] + 1

    # Vector to store the answer
    ans = []

    # Map to store the ratio
    mp = {}

    for i in range(n):
        # Find the gcd of pre0[i] and pre1[i]
        x = __gcd(pre0[i], pre1[i])

        # Converting the elements in
        # simplest form
        l = pre0[i] // x
        r = pre1[i] // x

        # Update the value in map
        if (f'[{l}, {r}]' in mp):
            mp[f'[{l}, {r}]'] += 1
        else:
            mp[f'[{l}, {r}]'] = 1

        # Store this in ans
        ans.append(mp[f'[{l}, {r}]'])

    # Return the ans vector
    for i in ans:
        print(i, end=" ")

# Driver Code
s = "001110"
equalSubstrings(s)

# This code is contributed by gfgking

C#

// C# program for the above approach
using System;
using System.Collections.Generic;

class GFG {

    // Function to retuan a prefix array of
    // equal partitions of the given string
    // consisting of 0s and 1s
    public static void equalSubstrings(string s)
    {

        // Length of the string
        int n = s.Length;

        // Prefix arrays for 0s and 1s
        int[] pre0 = new int[n];
        int[] pre1 = new int[n];

        // Arrays.fill(pre0, 0);
        // Arrays.fill(pre1, 0);

        // If character at index 0 is 0
        if (s[0] == '0') {
            pre0[0] = 1;
        }

        // If character at index 0 is 1
        else {
            pre1[0] = 1;
        }

        // Filling the prefix arrays
        for (int i = 1; i < n; i++) {
            if (s[i] == '0') {
                pre0[i] = pre0[i - 1] + 1;
                pre1[i] = pre1[i - 1];
            }
            else {
                pre0[i] = pre0[i - 1];
                pre1[i] = pre1[i - 1] + 1;
            }
        }

        // Vector to store the answer
        List<int> ans = new List<int>();

        // Map to store the ratio
        Dictionary<string, int> mp
            = new Dictionary<string, int>();

        for (int i = 0; i < n; i++) {

            // Find the gcd of pre0[i] and pre1[i]
            int x = __gcd(pre0[i], pre1[i]);

            // Converting the elements in
            // simplest form
            int l = pre0[i] / x, r = pre1[i] / x;

            string key = l + "," + r;

            // Update the value in map
            if (mp.ContainsKey(key))
                mp[key] += 1;
            else
                mp[key] = 1;

            // Store this in ans
            ans.Add(mp[key]);
        }

        // Return the ans vector
        foreach(int i in ans) Console.Write(i + " ");
    }

    public static int __gcd(int a, int b)
    {
        if (b == 0)
            return a;
        return __gcd(b, a % b);
    }

    // Driver Code
    public static void Main(string[] args)
    {
        string s = "001110";
        equalSubstrings(s);
    }
}

// This code is contributed by ukasp.

Javascript

<script>
    // JavaScript program for the above approach
    const __gcd = (a, b) => {
        if (b == 0) return a;
        return __gcd(b, a % b)
    }

    // Function to retuan a prefix array of
    // equal partitions of the given string
    // consisting of 0s and 1s

    const equalSubstrings = (s) => {
        // Length of the string
        let n = s.length;

        // Prefix arrays for 0s and 1s
        let pre0 = new Array(n).fill(0);
        let pre1 = new Array(n).fill(0);

        // If character at index 0 is 0
        if (s[0] == '0') {
            pre0[0] = 1;
        }

        // If character at index 0 is 1
        else {
            pre1[0] = 1;
        }

        // Filling the prefix arrays
        for (let i = 1; i < n; i++) {
            if (s[i] == '0') {
                pre0[i] = pre0[i - 1] + 1;
                pre1[i] = pre1[i - 1];
            }
            else {
                pre0[i] = pre0[i - 1];
                pre1[i] = pre1[i - 1] + 1;
            }
        }

        // Vector to store the answer
        ans = [];

        // Map to store the ratio
        mp = {};

        for (let i = 0; i < n; i++) {
            // Find the gcd of pre0[i] and pre1[i]
            let x = __gcd(pre0[i], pre1[i]);

            // Converting the elements in
            // simplest form
            let l = pre0[i] / x, r = pre1[i] / x;

            // Update the value in map
            if ([l, r] in mp) mp[[l, r]] += 1
            else mp[[l, r]] = 1

            // Store this in ans
            ans.push(mp[[l, r]]);
        }
        // Return the ans vector
        for (let i in ans)
            document.write(`${ans[i]} `);
    }

    // Driver Code

    let s = "001110";
    equalSubstrings(s);

    // This code is contributed by rakeshsahni

</script>

Output

1 2 1 1 1 2 

Time Complexity: O(nlogn)
Auxiliary Space: O(n)