在给定范围内具有相等元素的索引数

给定 N 个数字和 Q 个查询，每个查询由 L 和 R 组成，任务是找到这样的整数 i (L<=ii =A _i+1 。考虑基于 0 的索引。
例子 ：

Input : A = [1, 2, 2, 2, 3, 3, 4, 4, 4] 
        Q = 2 
        L = 1 R = 8 
        L = 0 R = 4 
Output : 5 
         2
Explanation: We have 5 index i which has 
Ai=Ai+1 in range [1, 8). We have 
2 indexes i which have Ai=Ai+1
in range [0, 4). 

Input :A = [3, 3, 4, 4] 
       Q = 2
       L = 0 R = 3
       L = 2 R = 3 
Output : 2 
         1

一种简单的方法是从 L 遍历到 R (Exclusive R) 并分别计算每个查询满足条件 A _i =A _i+1的索引 i 的数量。
以下是天真的方法的实现：

C++

// CPP program to count the number of indexes
// in range L R such that Ai = Ai+1
#include 
using namespace std;
 
// function that answers every query in O(r-l)
int answer_query(int a[], int n, int l, int r)
{
    // traverse from l to r and count
    // the required indexes
    int count = 0;
    for (int i = l; i < r; i++)
        if (a[i] == a[i + 1])
            count += 1;
 
    return count;
}
 
// Driver Code
int main()
{
    int a[] = { 1, 2, 2, 2, 3, 3, 4, 4, 4 };
    int n = sizeof(a) / sizeof(a[0]);
 
    // 1-st query
    int L, R;
    L = 1;
    R = 8;
 
    cout << answer_query(a, n, L, R) << endl;
 
    // 2nd query
    L = 0;
    R = 4;
    cout << answer_query(a, n, L, R) << endl;
    return 0;
}

Java

// Java program to count the number of
// indexes in range L R such that
// Ai = Ai+1
class GFG {
     
    // function that answers every query
    // in O(r-l)
    static int answer_query(int a[], int n,
                              int l, int r)
    {
         
        // traverse from l to r and count
        // the required indexes
        int count = 0;
        for (int i = l; i < r; i++)
            if (a[i] == a[i + 1])
                count += 1;
 
        return count;
    }
 
    // Driver Code
    public static void main(String[] args)
    {
        int a[] = {1, 2, 2, 2, 3, 3, 4, 4, 4};
        int n = a.length;
 
        // 1-st query
        int L, R;
        L = 1;
        R = 8;
 
        System.out.println(
                   answer_query(a, n, L, R));
 
        // 2nd query
        L = 0;
        R = 4;
        System.out.println(
                  answer_query(a, n, L, R));
    }
}
 
// This code is contribute by
// Smitha Dinesh Semwal

Python 3

# Python 3 program to count the
# number of indexes in range L R
# such that Ai = Ai + 1
 
# function that answers every
# query in O(r-l)
def answer_query(a, n, l, r):
 
    # traverse from l to r and
    # count the required indexes
    count = 0
    for i in range(l, r):
        if (a[i] == a[i + 1]):
            count += 1
 
    return count
 
# Driver Code
a = [1, 2, 2, 2, 3, 3, 4, 4, 4]
n = len(a)
 
# 1-st query
L = 1
R = 8
print(answer_query(a, n, L, R))
 
# 2nd query
L = 0
R = 4
print(answer_query(a, n, L, R))
 
# This code is contributed by
# Smitha Dinesh Semwal

C#

// C# program to count the number of
// indexes in range L R such that
// Ai = Ai+1
using System;
 
class GFG {
     
    // function that answers every query
    // in O(r-l)
    static int answer_query(int []a, int n,
                            int l, int r)
    {
         
        // traverse from l to r and count
        // the required indexes
        int count = 0;
        for (int i = l; i < r; i++)
            if (a[i] == a[i + 1])
                count += 1;
 
        return count;
    }
 
    // Driver Code
    public static void Main()
    {
        int []a = {1, 2, 2, 2, 3, 3,
                                4, 4, 4};
        int n = a.Length;
 
        // 1-st query
        int L, R;
        L = 1;
        R = 8;
 
        Console.WriteLine(
               answer_query(a, n, L, R));
 
        // 2nd query
        L = 0;
        R = 4;
        Console.WriteLine(
               answer_query(a, n, L, R));
    }
}
 
// This code is contribute by anuj_67.

Javascript

    // javascript program to count the number of
// indexes in range L R such that
// Ai = Ai+1
 
       
    // function that answers every query
    // in O(r-l)
     
    function answer_query(a,  n,  l, r)
    {
           
        // traverse from l to r and count
        // the required indexes
        var count = 0;
         
        for (var i = l; i < r; i++)
            if (a[i] == a[i + 1])
                count += 1;
   
        return count;
    }
   
    // Driver Code
 
        var a = [1, 2, 2, 2, 3, 3, 4, 4, 4]
        var n = a.length;
   
        // 1-st query
        var L, R;
        L = 1;
        R = 8;
   
        document.write(answer_query(a, n, L, R) + "<br>");
   
        // 2nd query
        L = 0;
        R = 4;
        document.write(answer_query(a, n, L, R));
     
 
   
 

输出：

5
2

时间复杂度：每个查询的 O(R – L)
高效的方法：我们可以在 O(1) 时间内回答查询。这个想法是预先计算一个前缀数组prefixans使得prefixans[i]存储从 0 到 i 的索引的总数，它服从给定的条件。 prefixans[R-1]-prefix[L-1]返回每个查询在 [L, r) 范围内的索引的总计数。
以下是有效方法的实施：

C++

// CPP program to count the number of indexes
// in range L R such that Ai=Ai+1
#include 
using namespace std;
const int N = 1000;
 
// array to store count of index from 0 to
// i that obey condition
int prefixans[N];
 
// precomputing prefixans[] array
int countIndex(int a[], int n)
{
    // traverse to compute the prefixans[] array
    for (int i = 0; i < n; i++) {
        if (a[i] == a[i + 1])
            prefixans[i] = 1;
 
        if (i != 0)
            prefixans[i] += prefixans[i - 1];
    }
}
 
// function that answers every query in O(1)
int answer_query(int l, int r)
{
    if (l == 0)
        return prefixans[r - 1];
    else
        return prefixans[r - 1] - prefixans[l - 1];
}
 
// Driver Code
int main()
{
    int a[] = { 1, 2, 2, 2, 3, 3, 4, 4, 4 };
    int n = sizeof(a) / sizeof(a[0]);
 
    // pre-computation
    countIndex(a, n);
 
    int L, R;
 
    // 1-st query
    L = 1;
    R = 8;
 
    cout << answer_query(L, R) << endl;
 
    // 2nd query
    L = 0;
    R = 4;
    cout << answer_query(L, R) << endl;
    return 0;
}

Java

// Java program to count
// the number of indexes
// in range L R such that
// Ai=Ai+1
 
class GFG {
     
public static int N = 1000;
 
// Array to store count
// of index from 0 to
// i that obey condition
static int prefixans[] = new int[1000];
 
// precomputing prefixans[] array
public static void countIndex(int a[], int n)
{
     
    // traverse to compute
    // the prefixans[] array
    for (int i = 0; i < n; i++) {
        if (i + 1 < n && a[i] == a[i + 1])
            prefixans[i] = 1;
 
        if (i != 0)
            prefixans[i] += prefixans[i - 1];
    }
}
 
// function that answers
// every query in O(1)
public static int answer_query(int l, int r)
{
    if (l == 0)
        return prefixans[r - 1];
    else
        return prefixans[r - 1] -
               prefixans[l - 1];
}
 
// Driver Code
public static void main(String args[])
{
    int a[] = {1, 2, 2, 2, 3, 3, 4, 4, 4};
    int n = 9;
 
    // pre-computation
    countIndex(a, n);
 
    int L, R;
 
    // 1-st query
    L = 1;
    R = 8;
 
    System.out.println(answer_query(L, R));
 
    // 2nd query
    L = 0;
    R = 4;
    System.out.println(answer_query(L, R));
}
}
 
// This code is contributed by Jaideep Pyne

Python3

# Python program to count
# the number of indexes in
# range L R such that Ai=Ai+1
N = 1000
 
# array to store count
# of index from 0 to
# i that obey condition
prefixans = [0] * N;
 
# precomputing
# prefixans[] array
def countIndex(a, n) :
 
    global N, prefixans
     
    # traverse to compute
    # the prefixans[] array
    for i in range(0, n - 1) :
 
        if (a[i] == a[i + 1]) :
            prefixans[i] = 1
 
        if (i != 0) :
            prefixans[i] = (prefixans[i] +
                            prefixans[i - 1])
 
# def that answers
# every query in O(1)
def answer_query(l, r) :
 
    global N, prefixans
    if (l == 0) :
        return prefixans[r - 1]
    else :
        return (prefixans[r - 1] -
                prefixans[l - 1])
 
# Driver Code
a = [1, 2, 2, 2,
     3, 3, 4, 4, 4]
n = len(a)
 
# pre-computation
countIndex(a, n)
 
# 1-st query
L = 1
R = 8
 
print (answer_query(L, R))
 
# 2nd query
L = 0
R = 4
print (answer_query(L, R))
 
# This code is contributed by
# Manish Shaw(manishshaw1)

C＃

// C# program to count the
// number of indexes in
// range L R such that Ai=Ai+1
using System;
 
class GFG
{
    static int N = 1000;
     
    // array to store count
    // of index from 0 to
    // i that obey condition
    static int []prefixans = new int[N];
     
    // precomputing prefixans[] array
    static void countIndex(int []a,
                           int n)
    {
        // traverse to compute
        // the prefixans[] array
        for (int i = 0; i < n - 1; i++)
        {
            if (a[i] == a[i + 1])
                prefixans[i] = 1;
     
            if (i != 0)
                prefixans[i] += prefixans[i - 1];
        }
    }
     
    // function that answers
    // every query in O(1)
    static int answer_query(int l, int r)
    {
        if (l == 0)
            return prefixans[r - 1];
        else
            return prefixans[r - 1] -
                   prefixans[l - 1];
    }
     
    // Driver Code
    static void Main()
    {
        int []a = new int[]{1, 2, 2, 2,
                            3, 3, 4, 4, 4};
        int n = a.Length;
         
        // pre-computation
        countIndex(a, n);
         
        int L, R;
     
        // 1-st query
        L = 1;
        R = 8;
     
        Console.WriteLine(answer_query(L, R));
     
        // 2nd query
        L = 0;
        R = 4;
        Console.WriteLine(answer_query(L, R));
    }
}
// This code is contributed by
// Manish Shaw(manishshaw1)

Javascript

 
// JavaScript program to count the number of indexes
// in range L R such that Ai=Ai+1
const N = 1000;
 
// array to store count of index from 0 to
// i that obey condition
let prefixans = new Uint8Array(N);
 
// precomputing prefixans[] array
function countIndex(a, n)
{
    // traverse to compute the prefixans[] array
    for (let i = 0; i < n; i++) {
        if (a[i] == a[i + 1])
            prefixans[i] = 1;
 
        if (i != 0)
            prefixans[i] += prefixans[i - 1];
    }
}
 
// function that answers every query in O(1)
function answer_query(l, r)
{
    if (l == 0)
        return prefixans[r - 1];
    else
        return prefixans[r - 1] - prefixans[l - 1];
}
 
// Driver Code
    let a = [ 1, 2, 2, 2, 3, 3, 4, 4, 4 ];
    let n = a.length;
 
    // pre-computation
    countIndex(a, n);
 
    let L, R;
 
    // 1-st query
    L = 1;
    R = 8;
 
    document.write(answer_query(L, R) + "<br>");
 
    // 2nd query
    L = 0;
    R = 4;
    document.write(answer_query(L, R) + "<br>");
 
 
// This code is contributed by Manoj.
 

输出：

5
2

时间复杂度：每个查询 O(1)，预计算 O(n)。