Count number of indices such that s[i] = s[i+1] : Range queries

Given a string str. Now for every query consisting of two integer L and R, the task is to find the number of indices such that str[i] = str[i+1] and L ≤ i, i+1 ≤ R.

Examples:

Input: str = “ggggggggggg”, query[] = {{1, 2}, {1, 5}}
Output: 1 4
The answer is 1 for first query and 4 for second query.
The condition is true for all indices except the last one in each query.

Input: str = “geeg”, query[] = {{0, 3}}
Output: 1
The condition is true only for i = 1.



Approach: Create a prefix array pref such that pref[i] holds the count of all the indices from 1 to i-1 such that str[i] = str[i+1]. Now for every query (L, R) the result will be pref[r] – pref[l].

Below is the implementation of the above approach:

C++

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// C++ program to find substring with  
#include <bits/stdc++.h>
using namespace std;
  
// Function to create prefix array
void preCompute(int n, string s, int pref[])
{
    pref[0] = 0;
    for (int i = 1; i < n; i++) {
        pref[i] = pref[i - 1];
        if (s[i - 1] == s[i])
            pref[i]++;
    }
}
  
// Function to return the result of the query
int query(int pref[], int l, int r)
{
    return pref[r] - pref[l];
}
  
// Driver Code
int main()
{
    string s = "ggggggg";
    int n = s.length();
  
    int pref[n];
    preCompute(n, s, pref);
  
    // Query 1
    int l = 1;
    int r = 2;
    cout << query(pref, l, r) << endl;
  
    // Query 2
    l = 1;
    r = 5;
    cout << query(pref, l, r) << endl;
  
    return 0;
}

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Java

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// Java program to find substring with 
  
import java.io.*;
  
class GFG {
  
// Function to create prefix array
static void preCompute(int n, String s, int pref[])
{
    pref[0] = 0;
    for (int i = 1; i < n; i++) {
        pref[i] = pref[i - 1];
        if (s.charAt(i - 1)== s.charAt(i))
            pref[i]++;
    }
}
  
// Function to return the result of the query
static int query(int pref[], int l, int r)
{
    return pref[r] - pref[l];
}
  
// Driver Code
  
    public static void main (String[] args) {
    String s = "ggggggg";
    int n = s.length();
  
    int pref[] = new int[n]; 
    preCompute(n, s, pref);
  
    // Query 1
    int l = 1;
    int r = 2;
    System.out.println( query(pref, l, r));
  
    // Query 2
    l = 1;
    r = 5;
    System.out.println(query(pref, l, r));
    }
}
// This code is contributed by inder_verma..

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Python3

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# Python3 program for the given approach
  
# Function to create prefix array 
def preCompute(n, s, pref): 
   
    for i in range(1,n):  
        pref[i] = pref[i - 1
        if s[i - 1] == s[i]: 
            pref[i] += 1
    
# Function to return the result of the query 
def query(pref, l, r): 
   
    return pref[r] - pref[l] 
    
if __name__ == "__main__":
  
    s = "ggggggg" 
    n = len(s) 
    
    pref = [0] * n
    preCompute(n, s, pref) 
    
    # Query 1 
    l = 1 
    r = 2 
    print(query(pref, l, r)) 
    
    # Query 2 
    l = 1 
    r = 5 
    print(query(pref, l, r)) 
      
# This code is contributed by Rituraj Jain

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C#

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// C# program to find substring with 
  
using System;
  
class GFG {
  
  
// Function to create prefix array
static void preCompute(int n, string s, int []pref)
{
    pref[0] = 0;
    for (int i = 1; i < n; i++) {
        pref[i] = pref[i - 1];
        if (s[i - 1]== s[i])
            pref[i]++;
    }
}
  
// Function to return the result of the query
static int query(int []pref, int l, int r)
{
    return pref[r] - pref[l];
}
  
// Driver Code
  
    public static void Main () {
    string s = "ggggggg";
    int n = s.Length;
  
    int []pref = new int[n]; 
    preCompute(n, s, pref);
  
    // Query 1
    int l = 1;
    int r = 2;
    Console.WriteLine( query(pref, l, r));
  
    // Query 2
    l = 1;
    r = 5;
    Console.WriteLine(query(pref, l, r));
    }
}
// This code is contributed by inder_verma..

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PHP

Output:

1
4


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Improved By : inderDuMCA, rituraj_jain, Ita_c