Count of substrings in a Binary String that contains more 1s than 0s

Given a binary string s, the task is to calculate the number of such substrings where the count of 1‘s is strictly greater than the count of 0‘s. 

Examples

Input: S = “110011”
Output: 11
Explanation: 
Substrings in which the count of 1’s is strictly greater than the count of 0’s are { S[0]}, {S[0], S[1]}, {S[0], S[2]}, {S[0], S[4]}, {S[0], S[5]}, {S[1], S[1]}, {S[1], S[5]}, {S[3], S[5]}, {S[4], S[4]}, {S[4], S[5]}, {S[5], S[5]}.

Input: S = “101”
Output: 3

Naive Approach: The simplest approach to solve the problem is to generate all substrings and count the number of 1s and 0s in each substring. Increase the count of those substrings that contain the count of 1s greater than the count of 0s. Finally, print the count obtained.
Time Complexity: O(N³)
Auxiliary Space: O(1)

Efficient Approach: The above approach can be optimized using Merge Sort Algorithm. Follow the steps below:

• Initialize an array, say nums[] of size n, where n is the length of the string.
• Traverse the string. If s[i]  == ‘1’, then store 1 in nums[i]. Otherwise, set nums[i] = -1.
• Update pref[] to store the prefix sum of the array nums[].
• Now, the problem reduces to counting the number of pairs(i, j) in the array pref[], where pref[i] < pref[j] and i < j, which is similar to counting inversions in an array from the rear side.
• Return the number of inversions of the prefix sum array as the final answer.

Below is the implementation of the above approach.

C++14

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

// Function to merge two partitions
// such that the merged array is sorted
void merge(vector<int>& v, int left,
           int mid, int right, int& inversions)
{
    vector<int> temp(right - left + 1);

    int i = left;
    int j = mid + 1;
    int k = 0;
    int cnt = 0;

    while (i <= mid && j <= right) {
        if (v[i] <= v[j]) {
            temp[k++] = v[i++];
        }
        else {
            // Counting inversions
            inversions += (mid - i + 1);

            temp[k++] = v[j++];
        }
    }

    while (i <= mid)
        temp[k++] = v[i++];

    while (j <= right)
        temp[k++] = v[j++];

    k = 0;
    for (int a = left; a <= right; a++) {
        v[a] = temp[k++];
    }
}

// Function to implement merge sort
void mergeSort(vector<int>& v, int left,
               int right, int& inversions)
{
    if (left < right) {
        int mid = (left + right) / 2;

        mergeSort(v, left, mid, inversions);
        mergeSort(v, mid + 1, right, inversions);
        merge(v, left, mid, right, inversions);
    }
}

// Function to calculate number of
// inversions in a given array
int CountInversions(vector<int>& v)
{
    int n = v.size();
    int inversions = 0;

    // Calculate the number of inversions
    mergeSort(v, 0, n - 1, inversions);

    // Return the number of inversions
    return inversions;
}

// Function to count the number of
// substrings that contains more 1s than 0s
int getSubsCount(string& input)
{
    int n = input.length();

    vector<int> nums(n);

    for (int i = 0; i < n; i++) {
        nums[i] = input[i] - '0';

        if (nums[i] == 0)
            nums[i] = -1;
    }

    // Stores the prefix sum array
    vector<int> pref(n);

    int sum = 0;

    for (int i = 0; i < n; i++) {
        sum += nums[i];
        pref[i] = sum;
    }

    int cnt = 0;

    // Stores the count of valid substrings
    for (int i = 0; i < n; i++) {
        if (pref[i] > 0)
            cnt++;
    }

    reverse(pref.begin(), pref.end());

    int inversions = CountInversions(pref);

    int ans = cnt + inversions;

    return ans;
}

// Driver Code
int main()
{

    // Given Input
    string input = "10011101";

    // Function Call
    int ans = getSubsCount(input);

    cout << ans << endl;

    return 0;
}

Java

import java.util.Arrays;

public class SubstringsCount {

    public static void merge(int[] arr, int left, int mid, int right, int[] inversions) {
        // Merge two sorted partitions and count inversions
        int[] temp = new int[right - left + 1];
        int i = left;
        int j = mid + 1;
        int k = 0;

        while (i <= mid && j <= right) {
            if (arr[i] <= arr[j]) {
                temp[k] = arr[i];
                k++;
                i++;
            } else {
                // Counting inversions when an element from the right partition is chosen
                inversions[0] += (mid - i + 1);
                temp[k] = arr[j];
                k++;
                j++;
            }
        }

        while (i <= mid) {
            temp[k] = arr[i];
            k++;
            i++;
        }

        while (j <= right) {
            temp[k] = arr[j];
            k++;
            j++;
        }

        k = 0;
        for (int a = left; a <= right; a++) {
            arr[a] = temp[k];
            k++;
        }
    }

    public static void mergeSort(int[] arr, int left, int right, int[] inversions) {
        // Recursive merge sort algorithm
        if (left < right) {
            int mid = (left + right) / 2;
            mergeSort(arr, left, mid, inversions);
            mergeSort(arr, mid + 1, right, inversions);
            merge(arr, left, mid, right, inversions);
        }
    }

    public static int countInversions(int[] arr) {
        // Calculate the number of inversions using the modified merge sort
        int[] inversions = {0};
        mergeSort(arr, 0, arr.length - 1, inversions);
        return inversions[0];
    }

    public static int getSubsCount(String inputStr) {
        int n = inputStr.length();
        int[] nums = new int[n];

        // Convert binary string to an array of 1s and -1s
        for (int i = 0; i < n; i++) {
            nums[i] = Character.getNumericValue(inputStr.charAt(i));
            if (nums[i] == 0) {
                nums[i] = -1;
            }
        }

        int[] pref = new int[n];
        int s = 0;

        // Calculate the prefix sum array
        for (int i = 0; i < n; i++) {
            s += nums[i];
            pref[i] = s;
        }

        // Count the number of valid substrings
        int cnt = (int) Arrays.stream(pref).filter(x -> x > 0).count();

        // Reverse the prefix sum array
        for (int i = 0; i < n / 2; i++) {
            int temp = pref[i];
            pref[i] = pref[n - 1 - i];
            pref[n - 1 - i] = temp;
        }

        // Calculate the number of inversions in the reversed prefix sum array
        int inversions = countInversions(pref);

        // Calculate the final result by adding counts of valid substrings and inversions
        return cnt + inversions;
    }

    // Driver Code
    public static void main(String[] args) {
        // Given Input
        String inputStr = "10011101";

        // Function Call
        int ans = getSubsCount(inputStr);

        // Print the result
        System.out.println(ans);
    }
}

Python3

def merge(arr, left, mid, right, inversions):
    # Merge two sorted partitions and count inversions
    temp = [0] * (right - left + 1)
    i = left
    j = mid + 1
    k = 0

    while i <= mid and j <= right:
        if arr[i] <= arr[j]:
            temp[k] = arr[i]
            k += 1
            i += 1
        else:
            # Counting inversions when an element from the right partition is chosen
            inversions[0] += (mid - i + 1)
            temp[k] = arr[j]
            k += 1
            j += 1

    while i <= mid:
        temp[k] = arr[i]
        k += 1
        i += 1

    while j <= right:
        temp[k] = arr[j]
        k += 1
        j += 1

    k = 0
    for a in range(left, right + 1):
        arr[a] = temp[k]
        k += 1


def merge_sort(arr, left, right, inversions):
    # Recursive merge sort algorithm
    if left < right:
        mid = (left + right) // 2
        merge_sort(arr, left, mid, inversions)
        merge_sort(arr, mid + 1, right, inversions)
        merge(arr, left, mid, right, inversions)


def count_inversions(arr):
    # Calculate the number of inversions using the modified merge sort
    n = len(arr)
    inversions = [0]
    merge_sort(arr, 0, n - 1, inversions)
    return inversions[0]


def get_subs_count(input_str):
    n = len(input_str)
    nums = [0] * n

    # Convert binary string to an array of 1s and -1s
    for i in range(n):
        nums[i] = int(input_str[i])
        if nums[i] == 0:
            nums[i] = -1

    pref = [0] * n
    s = 0

    # Calculate the prefix sum array
    for i in range(n):
        s += nums[i]
        pref[i] = s

    # Count the number of valid substrings
    cnt = sum(1 for x in pref if x > 0)

    # Reverse the prefix sum array
    pref.reverse()

    # Calculate the number of inversions in the reversed prefix sum array
    inversions = count_inversions(pref)

    # Calculate the final result by adding counts of valid substrings and inversions
    ans = cnt + inversions

    return ans


# Driver Code
if __name__ == "__main__":
    # Given Input
    input_str = "10011101"

    # Function Call
    ans = get_subs_count(input_str)

    # Print the result
    print(ans)

C#

using System;
using System.Linq;

public class SubstringsCount
{
    public static void Merge(int[] arr, int left, int mid, int right, int[] inversions)
    {
        // Merge two sorted partitions and count inversions
        int[] temp = new int[right - left + 1];
        int i = left;
        int j = mid + 1;
        int k = 0;

        while (i <= mid && j <= right)
        {
            if (arr[i] <= arr[j])
            {
                temp[k] = arr[i];
                k++;
                i++;
            }
            else
            {
                // Counting inversions when an element from the right partition is chosen
                inversions[0] += (mid - i + 1);
                temp[k] = arr[j];
                k++;
                j++;
            }
        }

        while (i <= mid)
        {
            temp[k] = arr[i];
            k++;
            i++;
        }

        while (j <= right)
        {
            temp[k] = arr[j];
            k++;
            j++;
        }

        k = 0;
        for (int a = left; a <= right; a++)
        {
            arr[a] = temp[k];
            k++;
        }
    }

    public static void MergeSort(int[] arr, int left, int right, int[] inversions)
    {
        // Recursive merge sort algorithm
        if (left < right)
        {
            int mid = (left + right) / 2;
            MergeSort(arr, left, mid, inversions);
            MergeSort(arr, mid + 1, right, inversions);
            Merge(arr, left, mid, right, inversions);
        }
    }

    public static int CountInversions(int[] arr)
    {
        // Calculate the number of inversions using the modified merge sort
        int[] inversions = { 0 };
        MergeSort(arr, 0, arr.Length - 1, inversions);
        return inversions[0];
    }

    public static int GetSubsCount(string inputStr)
    {
        int n = inputStr.Length;
        int[] nums = new int[n];

        // Convert binary string to an array of 1s and -1s
        for (int i = 0; i < n; i++)
        {
            nums[i] = int.Parse(inputStr[i].ToString());
            if (nums[i] == 0)
            {
                nums[i] = -1;
            }
        }

        int[] pref = new int[n];
        int s = 0;

        // Calculate the prefix sum array
        for (int i = 0; i < n; i++)
        {
            s += nums[i];
            pref[i] = s;
        }

        // Count the number of valid substrings
        int cnt = pref.Count(x => x > 0);

        // Reverse the prefix sum array
        Array.Reverse(pref);

        // Calculate the number of inversions in the reversed prefix sum array
        int inversions = CountInversions(pref);

        // Calculate the final result by adding counts of valid substrings and inversions
        return cnt + inversions;
    }

    // Driver Code
    public static void Main(string[] args)
    {
        // Given Input
        string inputStr = "10011101";

        // Function Call
        int ans = GetSubsCount(inputStr);

        // Print the result
        Console.WriteLine(ans);
    }
}

Javascript

class SubstringsCount {
    static merge(arr, left, mid, right, inversions) {
        // Merge two sorted partitions and count inversions
        let temp = new Array(right - left + 1);
        let i = left;
        let j = mid + 1;
        let k = 0;

        while (i <= mid && j <= right) {
            if (arr[i] <= arr[j]) {
                temp[k] = arr[i];
                k++;
                i++;
            } else {
                // Counting inversions when an element from the right partition is chosen
                inversions[0] += (mid - i + 1);
                temp[k] = arr[j];
                k++;
                j++;
            }
        }

        while (i <= mid) {
            temp[k] = arr[i];
            k++;
            i++;
        }

        while (j <= right) {
            temp[k] = arr[j];
            k++;
            j++;
        }

        k = 0;
        for (let a = left; a <= right; a++) {
            arr[a] = temp[k];
            k++;
        }
    }

    static mergeSort(arr, left, right, inversions) {
        // Recursive merge sort algorithm
        if (left < right) {
            let mid = Math.floor((left + right) / 2);
            this.mergeSort(arr, left, mid, inversions);
            this.mergeSort(arr, mid + 1, right, inversions);
            this.merge(arr, left, mid, right, inversions);
        }
    }

    static countInversions(arr) {
        // Calculate the number of inversions using the modified merge sort
        let inversions = [0];
        this.mergeSort(arr, 0, arr.length - 1, inversions);
        return inversions[0];
    }

    static getSubsCount(inputStr) {
        let n = inputStr.length;
        let nums = new Array(n);

        // Convert binary string to an array of 1s and -1s
        for (let i = 0; i < n; i++) {
            nums[i] = parseInt(inputStr[i]);
            if (nums[i] === 0) {
                nums[i] = -1;
            }
        }

        let pref = new Array(n);
        let s = 0;

        // Calculate the prefix sum array
        for (let i = 0; i < n; i++) {
            s += nums[i];
            pref[i] = s;
        }

        // Count the number of valid substrings
        let cnt = pref.filter(x => x > 0).length;

        // Reverse the prefix sum array
        pref.reverse();

        // Calculate the number of inversions in the reversed prefix sum array
        let inversions = this.countInversions(pref);

        // Calculate the final result by adding counts of valid substrings and inversions
        return cnt + inversions;
    }
}

// Driver Code
let inputStr = "10011101";
let ans = SubstringsCount.getSubsCount(inputStr);
console.log(ans);

Output

23

Time Complexity: O(NlogN)
Auxiliary Space: O(N)

Segment Tree Approach

The below code uses segment tree. This approach is also O(NlogN) and can be seen yet as another approach.

The subarray [i,j] should have atleast sum 1, if we consider -1 for ‘0’ and 1 for ‘1’.

Mathematically,
      prefix[j] – prefix[i] >=1
=> prefix[i] <=  prefix[j]-1

Using segment tree, count all prefix[i] which are less than or equal to for 1 – prefix[j].

Now, here is one more thing. What if the string has a negative prefix sum, i.e S=”001″. The prefix sum at index 1 is -2. You cannot have negative indices / ranges in segment tree.

For that we do index shifting. Simply saying, its like shifting the origin by some value.

This helps us in overcoming negative ranges in segment tree. If N<=100000, So the highest prefix sum can be 100000 and the lowest prefix sum can be -100000. You cannot have negative range/index for segment tree. So -100000 won’t be possible.

If we add 100000 from our side manually. The new highest and lowest becomes 200000 and 0 respectively. As you can see, the relative distance still is the same, but there is no negative and hence can be used in segment tree.

For any string S, the shift value will be the size of the string.

Follow the steps below to implement the above idea:

1) Make a vector “tree”, and resize it according to the size of the input. Hereby, 4*(2*n)+1. 2*n because of the shift phenomenon as explained before. For any string S, the shift value will be the size of the string.2) Use the generic code for update and query function in segment tree.

3) Update the current running sum i.e. 0+shiftValue in the segment tree. Zero sum because there is no prefix sum as of now.

4) Iterate from i=0 to n. For S[i]=’1′ ,add 1 and for S[i]=’0′ , add -1. As discussed earlier, we will add shift value to this sum and search for the number of indices where prefix sum was less than or equal to (currSum + shift) – 1. For this, we search in the range 0 to (currSum + shift) – 1, as the lowest possible value can be 0, because of shifting and the highest value we can have is (currSum + shift) – 1. Add the number of indices to the answer.

5) Update the current running prefix sum with its shift value added to it, i.e. update currSum+shift in the segment tree.

6) Return the answer after the loop is terminated.

C++

// Author - RainX (Abhijit Roy, NIT AGARTALA)

#include <bits/stdc++.h>
using namespace std;

vector<int> tree;

void update(int start, int end, int parent, long long index){
   if (start > end) {
       return;
   }
   if (start == end) {
       tree[parent]++;
       return;
   }
   int mid = (start + end) / 2;
   if (index > mid) {
       update(mid + 1, end, 2 * parent + 2, index);
   }
   else {
       update(start, mid, 2 * parent + 1, index);
   }
   tree[parent] = tree[2 * parent + 1] + tree[2 * parent + 2];
}

int query(int start, int end, int parent, int qstart, int qend){
   if (qstart > end || qend < start) {
       return 0;
   }
   if (qstart <= start && qend >= end) {
       return tree[parent];
   }
   int mid = (start + end) / 2;
   int L = query(start, mid, 2 * parent + 1, qstart, qend);
   int R = query(mid + 1, end, 2 * parent + 2, qstart, qend);
   return L + R;
}

int getSubsCount(string &S){
   int n = S.size();
   tree.resize(4 * 2 * n + 1, 0);

   int shift = n;
   long long currSum = 0;
   long long cnt = 0;

   update(0, 2 * n, 0, 0 + shift);

   for (int i = 0; i < n; i++) {
       currSum += (S[i] == '1' ? 1 : -1);

       /*    prefix[j]-prefix[i]>=1
          => prefix[i]<=prefix[j]-1
       */

       int lessThan = (currSum + shift) - 1;
       cnt += query(0, 2 * n, 0, 0, lessThan);

       update(0, 2 * n, 0, currSum + shift);
   }
   return cnt;
}

int main(){
   string input = "10011101";

   int ans = getSubsCount(input);

   cout << ans << endl;

   return 0;
}

// Author - RainX (Abhijit Roy, NIT AGARTALA,2023)

Java

import java.util.*;

class Main {
    static List<Integer> tree;

    static void update(int start, int end, int parent, long index) {
        if (start > end) {
            return;
        }
        if (start == end) {
            tree.set(parent, tree.get(parent) + 1);
            return;
        }
        int mid = (start + end) / 2;
        if (index > mid) {
            update(mid + 1, end, 2 * parent + 2, index);
        }
        else {
            update(start, mid, 2 * parent + 1, index);
        }
        tree.set(parent, tree.get(2 * parent + 1) + tree.get(2 * parent + 2));
    }

    static int query(int start, int end, int parent, int qstart, int qend) {
        if (qstart > end || qend < start) {
            return 0;
        }
        if (qstart <= start && qend >= end) {
            return tree.get(parent);
        }
        int mid = (start + end) / 2;
        int L = query(start, mid, 2 * parent + 1, qstart, qend);
        int R = query(mid + 1, end, 2 * parent + 2, qstart, qend);
        return L + R;
    }

    static int getSubsCount(String S) {
        int n = S.length();
        tree = new ArrayList<Integer>(Collections.nCopies(4 * 2 * n + 1, 0));

        int shift = n;
        long currSum = 0;
        long cnt = 0;

        update(0, 2 * n, 0, 0 + shift);

        for (int i = 0; i < n; i++) {
            currSum += (S.charAt(i) == '1' ? 1 : -1);

            /*    prefix[j]-prefix[i]>=1
                => prefix[i]<=1-prefix[j]
            */

            int lessThan = (int) (currSum + shift) - 1;
            cnt += query(0, 2 * n, 0, 0, lessThan);

            update(0, 2 * n, 0, currSum + shift);
        }
        return (int) cnt;
    }

    public static void main(String[] args) {
        String input = "10011101";
        int ans = getSubsCount(input);
        System.out.println(ans);
    }
}

Python3

from typing import List


def update(start: int, end: int, parent: int, index: int) -> None:
    if start > end:
        return
    if start == end:
        tree[parent] += 1
        return
    mid = (start + end) // 2
    if index > mid:
        update(mid + 1, end, 2 * parent + 2, index)
    else:
        update(start, mid, 2 * parent + 1, index)
    tree[parent] = tree[2 * parent + 1] + tree[2 * parent + 2]


def query(start: int, end: int, parent: int, qstart: int, qend: int) -> int:
    if qstart > end or qend < start:
        return 0
    if qstart <= start and qend >= end:
        return tree[parent]
    mid = (start + end) // 2
    L = query(start, mid, 2 * parent + 1, qstart, qend)
    R = query(mid + 1, end, 2 * parent + 2, qstart, qend)
    return L + R


def getSubsCount(S: str) -> int:
    n = len(S)
    global tree
    tree = [0] * (4 * 2 * n + 1)

    shift = n
    currSum = 0
    cnt = 0

    update(0, 2 * n, 0, 0 + shift)

    for i in range(n):
        currSum += 1 if S[i] == '1' else -1

        """
        prefix[j] - prefix[i] >= 1
        => prefix[i] <= 1 - prefix[j]
        """

        lessThan = int(currSum + shift) - 1
        cnt += query(0, 2 * n, 0, 0, lessThan)

        update(0, 2 * n, 0, int(currSum + shift))

    return cnt


if __name__ == '__main__':
    input_str = "10011101"
    ans = getSubsCount(input_str)
    print(ans)

C#

using System;
using System.Collections.Generic;

public class MainClass {
    static List<int> tree;

    static void update(int start, int end, int parent, long index) {
        if (start > end) {
            return;
        }
        if (start == end) {
            tree[parent] += 1;
            return;
        }
        int mid = (start + end) / 2;
        if (index > mid) {
            update(mid + 1, end, 2 * parent + 2, index);
        }
        else {
            update(start, mid, 2 * parent + 1, index);
        }
        tree[parent] = tree[2 * parent + 1] + tree[2 * parent + 2];
    }

    static int query(int start, int end, int parent, int qstart, int qend) {
        if (qstart > end || qend < start) {
            return 0;
        }
        if (qstart <= start && qend >= end) {
            return tree[parent];
        }
        int mid = (start + end) / 2;
        int L = query(start, mid, 2 * parent + 1, qstart, qend);
        int R = query(mid + 1, end, 2 * parent + 2, qstart, qend);
        return L + R;
    }

    static int getSubsCount(string S) {
        int n = S.Length;
        tree = new List<int>(new int[4 * 2 * n + 1]);

        int shift = n;
        long currSum = 0;
        long cnt = 0;

        update(0, 2 * n, 0, 0 + shift);

        for (int i = 0; i < n; i++) {
            currSum += (S[i] == '1' ? 1 : -1);

            /*    prefix[j]-prefix[i]>=1
                => prefix[i]<=1-prefix[j]
            */

            int lessThan = (int) (currSum + shift) - 1;
            cnt += query(0, 2 * n, 0, 0, lessThan);

            update(0, 2 * n, 0, currSum + shift);
        }
        return (int) cnt;
    }

    public static void Main() {
        string input = "10011101";
         int ans = getSubsCount(input);
       Console.WriteLine(ans);
    }
}

Javascript

function update(start, end, parent, index, tree) {
    if (start > end) {
        return;
    }
    if (start == end) {
        tree[parent] += 1;
        return;
    }
    const mid = Math.floor((start + end) / 2);
    if (index > mid) {
        update(mid + 1, end, 2 * parent + 2, index, tree);
    }
    else {
        update(start, mid, 2 * parent + 1, index, tree);
    }
    tree[parent] = tree[2 * parent + 1] + tree[2 * parent + 2];
}

function query(start, end, parent, qstart, qend, tree) {
    if (qstart > end || qend < start) {
        return 0;
    }
    if (qstart <= start && qend >= end) {
        return tree[parent];
    }
    const mid = Math.floor((start + end) / 2);
    const L = query(start, mid, 2 * parent + 1, qstart, qend, tree);
    const R = query(mid + 1, end, 2 * parent + 2, qstart, qend, tree);
    return L + R;
}

function getSubsCount(S) {
    const n = S.length;
    const tree = Array.from({length: 4 * 2 * n + 1}, () => 0);

    const shift = n;
    let currSum = 0;
    let cnt = 0;

    update(0, 2 * n, 0, 0 + shift, tree);

    for (let i = 0; i < n; i++) {
        currSum += (S.charAt(i) == '1' ? 1 : -1);

        /*    prefix[j]-prefix[i]>=1
            => prefix[i]<=1-prefix[j]
        */

        const lessThan = Math.floor(currSum + shift) - 1;
        cnt += query(0, 2 * n, 0, 0, lessThan, tree);

        update(0, 2 * n, 0, currSum + shift, tree);
    }
    return cnt;
}

const input = "10011101";
const ans = getSubsCount(input);
console.log(ans);

Output

23

Time Complexity – O(NlogN) , O(logN) for query search in segment tree

Auxiliary Space – O(4*N)  ? O(N)