# 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.

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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++

 `// C++ program to find substring with   ` `#include ` `using` `namespace` `std; ` ` `  `// Function to create prefix array ` `void` `preCompute(``int` `n, string s, ``int` `pref[]) ` `{ ` `    ``pref = 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; ` `} `

## Java

 `// 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.. `

## Python3

 `# 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 `

## C#

 `// 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; ` `    ``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.. `

## PHP

 ` `

Output:

```1
4
```

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