Number of substrings with length divisible by the number of 1’s in it

Last Updated : 19 Dec, 2022

Given a binary string S consisting of only 0’s and 1’s. Count the number of substrings of this string such that the length of the substring is divisible by the number of 1’s in the substring.

Examples:

Input: S = “01010”
Output: 10

Input: S = “1111100000”
Output: 25

Naive Approach:

• Iterate through all the substrings and for each substring count the number of 1’s. If the length of the substring is divisible by the count of 1’s, then increase the count. This approach will take O(N3) time.
• To make the solution more efficient, instead of traversing the entire substring to count the number of 1’s in it, we can use Prefix Sum Array.

number of 1’s in the substring [l: r] = prefix_sum[r] – prefix_sum[l – 1]

Below is the implementation of the above approach.

C++

 // C++ program to count number of // substring under given condition #include using namespace std; // Function return count of // such substringint countOfSubstrings(string s){    int n = s.length();     int prefix_sum[n] = { 0 };     // Mark 1 at those indices     // where '1' appears    for (int i = 0; i < n; i++) {        if (s[i] == '1')            prefix_sum[i] = 1;    }     // Take prefix sum    for (int i = 1; i < n; i++)        prefix_sum[i] += prefix_sum[i - 1];     int answer = 0;     // Iterate through all the     // substrings    for (int i = 0; i < n; i++) {        for (int j = i; j < n; j++) {            int countOfOnes = prefix_sum[j]                - (i - 1 >= 0 ?                    prefix_sum[i - 1] : 0);            int length = j - i + 1;             if (countOfOnes > 0 &&                 length % countOfOnes == 0)                answer++;        }    }    return answer;} // Driver Codeint main(){    string S = "1111100000";     cout << countOfSubstrings(S);     return 0;}

Java

 // Java program to count number of // subString under given condition  import java.util.*; class GFG{ // Function return count of // such substringstatic int countOfSubStrings(String s){    int n = s.length();     int []prefix_sum = new int[n];     // Mark 1 at those indices     // where '1' appears    for (int i = 0; i < n; i++) {        if (s.charAt(i) == '1')            prefix_sum[i] = 1;    }         // Take prefix sum    for (int i = 1; i < n; i++)        prefix_sum[i] += prefix_sum[i - 1];     int answer = 0;         // Iterate through all the     // subStrings    for (int i = 0; i < n; i++) {        for (int j = i; j < n; j++) {            int countOfOnes = prefix_sum[j]                - (i - 1 >= 0 ?                 prefix_sum[i - 1] : 0);            int length = j - i + 1;             if (countOfOnes > 0 &&                 length % countOfOnes == 0)                answer++;        }    }    return answer;} // Driver Codepublic static void main(String[] args){    String S = "1111100000";     System.out.print(countOfSubStrings(S)); }} // This code contributed by sapnasingh4991

Python3

 # Python3 program to count number of # substring under given condition  # Function return count of # such substringdef countOfSubstrings(s):    n = len(s)    prefix_sum = [0 for i in range(n)]     # Mark 1 at those indices     # where '1' appears    for i in range(n):        if (s[i] == '1'):            prefix_sum[i] = 1     # Take prefix sum    for i in range(1, n, 1):        prefix_sum[i] += prefix_sum[i - 1]     answer = 0     # Iterate through all the     # substrings    for i in range(n):        for j in range(i, n, 1):            if (i - 1 >= 0):                countOfOnes = prefix_sum[j]- prefix_sum[i - 1]            else:                countOfOnes = prefix_sum[j]            length = j - i + 1             if (countOfOnes > 0 and length % countOfOnes == 0):                answer += 1     return answer # Driver Codeif __name__ == '__main__':    S = "1111100000"     print(countOfSubstrings(S)) # This code is contributed by Bhupendra_Singh

C#

 // C# program to count number of // subString under given condition using System;  class GFG{  // Function return count of // such substring static int countOfSubStrings(String s) {     int n = s.Length;     int []prefix_sum = new int[n];      // Mark 1 at those indices     // where '1' appears     for(int i = 0; i < n; i++)     {         if (s[i] == '1')             prefix_sum[i] = 1;     }          // Take prefix sum     for(int i = 1; i < n; i++)         prefix_sum[i] += prefix_sum[i - 1];      int answer = 0;          // Iterate through all the     // subStrings     for(int i = 0; i < n; i++)     {         for(int j = i; j < n; j++)         {             int countOfOnes = prefix_sum[j] -                                   (i - 1 >= 0 ?                               prefix_sum[i - 1] : 0);             int length = j - i + 1;                  if (countOfOnes > 0 &&                 length % countOfOnes == 0)             answer++;         }     }     return answer; }  // Driver Code public static void Main(String[] args) {     String S = "1111100000";     Console.Write(countOfSubStrings(S)); } }  // This code is contributed by Rohit_ranjan

Javascript

 

Output:
25

Time Complexity: O(N2)

Auxiliary Space: O(N)

Efficient Approach:

• Let’s use the prefix sum array as discussed in the previous approach. Let’s take a substring [l: r] which follows the condition given, we can say that:

r – l + 1 = k * (prefix_sum[r] – prefix_sum[l-1]), where k is some integer

• This can also be written as:

r – k * prefix_sum[r] = l – 1 – k * prefix_sum[l-1]

Observing this equation, we can create another array of the B[i] = i – k * prefix_sum[i], for 0 <= k < n. So, the problem now is to calculate the number of pairs of equal integers in B for each k.

• Let’s take a fixed number x. If k > x, then as r – l + 1 <= n ( for any pair of l, r ), we get the following inequality :

prefix_sum[r] – prefix_sum[l-1] = (r – l + 1)/k <= n/x

• This goes on to show that the number of ones and k is not large in substrings satisfying the given conditions. (this is just for estimation of efficiency). Now for k <= x, we need to calculate the number of pairs of equal integers. The following inequality is satisfied here:

(-1) *n*x <= i – k – prefix_sum[i] <= n

• So, we can put all numbers into one array for each k independently and find the number of equal integers.
• The next step would be to fix values of l and iterate on the number of 1’s in the string, and for each value get the bounds for r. For a given count of the number of 1’s, we can evaluate which remainder will give r. The problem now becomes is to calculate the number of integers on some segments, which gives some fixed remainder (rem) when divided by some fixed integer. This calculation can be done in O(1). Also, note that it’s important to count only such r values for which k > x.

Below is the implementation of the above approach.

C++

 // C++ program to count number of // substring under given condition #include using namespace std; // Function return count of such // substringint countOfSubstrings(string s){     int n = s.length();     // Selection of adequate x value    int x = sqrt(n);     // Store where 1's are located    vector<int> ones;     for (int i = 0; i < n; i++) {        if (s[i] == '1')            ones.push_back(i);    }     // If there are no ones,    // then answer is 0    if (ones.size() == 0)        return 0;     // For ease of implementation    ones.push_back(n);     // Count storage    vector<int> totCount(n * x + n);     int sum = 0;     // Iterate on all k values less     // than fixed x    for (int k = 0; k <= x; k++) {        // Keeps a count of 1's occurred        // during string traversal        int now = 0;        totCount[k * n]++;         // Iterate on string and modify         // the totCount        for (int j = 1; j <= n; j++) {            // If this character is 1            if (s[j - 1] == '1')                now++;            int index = j - k * now;             // Add to the final sum/count            sum += totCount[index + k * n];                         // Increase totCount at exterior             // position            totCount[index + k * n]++;        }        now = 0;        totCount[k * n]--;         for (int j = 1; j <= n; j++) {            if (s[j - 1] == '1')                now++;            int index = j - k * now;             // Reduce totCount at index + k*n            totCount[index + k * n]--;        }    }     // Slightly modified prefix sum storage    int prefix_sum[n];    memset(prefix_sum, -1, sizeof(prefix_sum));     // Number of 1's till i-1    int cnt = 0;     for (int i = 0; i < n; i++) {        prefix_sum[i] = cnt;        if (s[i] == '1')            cnt++;    }     // Traversing over string considering    // each position and finding bounds     // and count using the inequalities    for (int k = 0; k < n; k++)    {        for (int j = 1; j <= (n / x)              && prefix_sum[k] + j <= cnt; j++)        {            // Calculating bounds for l and r            int l = ones[prefix_sum[k] + j - 1]                    - k + 1;                         int r = ones[prefix_sum[k] + j] - k;                         l = max(l, j * (x + 1));             // If valid then add to answer            if (l <= r) {                sum += r / j - (l - 1) / j;            }        }    }     return sum;}int main(){    string S = "1111100000";     cout << countOfSubstrings(S);     return 0;}

Java

 // Java program to count number of // subString under given condition import java.util.*; class GFG{ // Function return count of such // substringstatic int countOfSubStrings(String s){    int n = s.length();     // Selection of adequate x value    int x = (int)Math.sqrt(n);     // Store where 1's are located    Vector ones = new Vector();     for(int i = 0; i < n; i++)     {        if (s.charAt(i) == '1')            ones.add(i);    }     // If there are no ones,    // then answer is 0    if (ones.size() == 0)        return 0;     // For ease of implementation    ones.add(n);     // Count storage    int []totCount = new int[n * x + n];     int sum = 0;     // Iterate on all k values less     // than fixed x    for(int k = 0; k <= x; k++)    {                 // Keeps a count of 1's occurred        // during String traversal        int now = 0;        totCount[k * n]++;         // Iterate on String and modify         // the totCount        for(int j = 1; j <= n; j++)        {                         // If this character is 1            if (s.charAt(j - 1) == '1')                now++;                             int index = j - k * now;             // Add to the final sum/count            sum += totCount[index + k * n];                         // Increase totCount at exterior             // position            totCount[index + k * n]++;        }        now = 0;        totCount[k * n]--;         for(int j = 1; j <= n; j++)        {            if (s.charAt(j - 1) == '1')                now++;                             int index = j - k * now;             // Reduce totCount at index + k*n            totCount[index + k * n]--;        }    }     // Slightly modified prefix sum storage    int []prefix_sum = new int[n];    Arrays.fill(prefix_sum, -1);     // Number of 1's till i-1    int cnt = 0;     for(int i = 0; i < n; i++)    {        prefix_sum[i] = cnt;        if (s.charAt(i) == '1')            cnt++;    }     // Traversing over String considering    // each position and finding bounds     // and count using the inequalities    for(int k = 0; k < n; k++)    {        for(int j = 1; j <= (n / x) &&             prefix_sum[k] + j <= cnt; j++)        {                         // Calculating bounds for l and r            int l = ones.get(prefix_sum[k] + j - 1)-                                              k + 1;                         int r = ones.get(prefix_sum[k] + j) - k;            l = Math.max(l, j * (x + 1));             // If valid then add to answer            if (l <= r)            {                sum += r / j - (l - 1) / j;            }        }    }    return sum;} // Driver codepublic static void main(String[] args){    String S = "1111100000";     System.out.print(countOfSubStrings(S));}} // This code is contributed by Amit Katiyar

Python3

 # Python3 program to count number of # substring under given condition import math # Function return count of such # substringdef countOfSubstrings(s):     n = len(s)     # Selection of adequate x value    x = int(math.sqrt(n))     # Store where 1's are located    ones = []     for i in range (n):        if (s[i] == '1'):            ones.append(i)     # If there are no ones,    # then answer is 0    if (len(ones) == 0):        return 0     # For ease of implementation    ones.append(n)     # Count storage    totCount = [0] * (n * x + n)     sum = 0     # Iterate on all k values less     # than fixed x    for k in range(x + 1):                 # Keeps a count of 1's occurred        # during string traversal        now = 0        totCount[k * n] += 1         # Iterate on string and modify         # the totCount        for j in range(1, n + 1):                         # If this character is 1            if (s[j - 1] == '1'):                now += 1            index = j - k * now             # Add to the final sum/count            sum += totCount[index + k * n]                         # Increase totCount at exterior             # position            totCount[index + k * n] += 1                 now = 0        totCount[k * n] -= 1         for j in range(1, n + 1):            if (s[j - 1] == '1'):                now += 1            index = j - k * now             # Reduce totCount at index + k*n            totCount[index + k * n] -= 1     # Slightly modified prefix sum storage    prefix_sum = [-1] * n     # Number of 1's till i-1    cnt = 0     for i in range(n):        prefix_sum[i] = cnt        if (s[i] == '1'):            cnt += 1         # Traversing over string considering    # each position and finding bounds     # and count using the inequalities    for k in range(n):        j = 1        while (j <= (n // x) and               prefix_sum[k] + j <= cnt):                     # Calculating bounds for l and r            l = (ones[prefix_sum[k] + j - 1] -                                      k + 1)                         r = ones[prefix_sum[k] + j] - k            l = max(l, j * (x + 1))             # If valid then add to answer            if (l <= r):                sum += r // j - (l - 1) // j                         j += 1     return sum # Driver codeif __name__ == "__main__":         S = "1111100000"     print (countOfSubstrings(S)) # This code is contributed by chitranayal

C#

 // C# program to count number of // subString under given condition using System;using System.Collections.Generic;  class GFG{  // Function return count of such // substringstatic int countOfSubStrings(String s){    int n = s.Length;      // Selection of adequate x value    int x = (int)Math.Sqrt(n);      // Store where 1's are located    List<int> ones = new List<int>();      for(int i = 0; i < n; i++)     {        if (s[i] == '1')            ones.Add(i);    }      // If there are no ones,    // then answer is 0    if (ones.Count == 0)        return 0;      // For ease of implementation    ones.Add(n);      // Count storage    int []totCount = new int[n * x + n];      int sum = 0;      // Iterate on all k values less     // than fixed x    for(int k = 0; k <= x; k++)    {                  // Keeps a count of 1's occurred        // during String traversal        int now = 0;        totCount[k * n]++;          // Iterate on String and modify         // the totCount        for(int j = 1; j <= n; j++)        {                          // If this character is 1            if (s[j-1] == '1')                now++;                              int index = j - k * now;              // Add to the readonly sum/count            sum += totCount[index + k * n];                          // Increase totCount at exterior             // position            totCount[index + k * n]++;        }        now = 0;        totCount[k * n]--;          for(int j = 1; j <= n; j++)        {            if (s[j-1] == '1')                now++;                              int index = j - k * now;              // Reduce totCount at index + k*n            totCount[index + k * n]--;        }    }      // Slightly modified prefix sum storage    int []prefix_sum = new int[n];    for(int i = 0; i < n; i++)        prefix_sum [i]= -1;      // Number of 1's till i-1    int cnt = 0;      for(int i = 0; i < n; i++)    {        prefix_sum[i] = cnt;        if (s[i] == '1')            cnt++;    }      // Traversing over String considering    // each position and finding bounds     // and count using the inequalities    for(int k = 0; k < n; k++)    {        for(int j = 1; j <= (n / x) &&             prefix_sum[k] + j <= cnt; j++)        {                          // Calculating bounds for l and r            int l = ones[prefix_sum[k] + j - 1]-                                          k + 1;                          int r = ones[prefix_sum[k] + j] - k;            l = Math.Max(l, j * (x + 1));              // If valid then add to answer            if (l <= r)            {                sum += r / j - (l - 1) / j;            }        }    }    return sum;}  // Driver codepublic static void Main(String[] args){    String S = "1111100000";      Console.Write(countOfSubStrings(S));}}  // This code is contributed by Amit Katiyar

Javascript

 

Output:
25

Time Complexity :

Auxiliary Space: O(N3/2)

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