没有更新的给定数组中的修改范围总和

给定一个大小为N的数组arr[]包含从 1 到 N 的任意顺序的不同数字，任务是根据以下规则在该数组中执行修改后的范围和。
对于数组arr 中的每个索引“ i ”：

• 范围 ‘ L ‘ 的起始索引选择为i + 1
• 范围 ‘ R ‘ 的结束索引选择为：
  • 分钟(arr[i]，N-1) ；如果 arr[i] > i
  • max(i+1, arr[i]) ;如果 arr[i] < i+1
• 对于更新，范围 arr[L] 到 arr[R] 中的值增加 1。
• 使用输入数组而不是更新的数组找到范围

例子：

Input: arr[] = {4, 1, 3, 2} 
Output: 4 2 5 4 
Explanation: 
For i = 0 -> Element in input array = 4. Therefore L = 1, and R = min(4, N-1) = 3. Hence, all the elements from arr[1] to arr[3] are incremented by 1. The elements after update operation are {4, 2, 4, 3}. 
For i = 1 -> Element in input array = 1. Therefore L = 2, and R = max(1, i+1) = 2. Hence, all the elements from arr[2] to arr[2] are incremented by 1. The elements after update operation are {4, 2, 5, 3}. 
For i = 2 -> Element in input array = 3. Therefore L = 3, and R = min(3, N-1) = 3. Hence, all the elements from arr[3] to arr[3] are incremented by 1. The elements after update operation are {4, 2, 5, 4}. 
For i = 3 -> The array is unaffected. Therefore the elements after update operation are {4, 2, 5, 4}. 
The resulting array is {4, 2, 5, 4}.
Input: arr[] = {2, 1} 
Output: {2, 2} 
Explanation: 
The first element is 2. So arr[1] gets incremented by 1. Hence, the resulting array is {2, 2}. 
 

朴素的方法：朴素的方法是为每个元素运行一个循环，并将从 arr[i+1] 到 arr[min(i+arr[i], N-1)] 的所有值增加 1。这种方法是O(N ² ) 。
高效的方法：这个问题可以通过使用 O(N) 的额外空间在 O(N) 中解决。这个想法是使用前缀和数组的概念。按照以下步骤计算答案：

• 声明了一个大小为N + 1的数组b[]并且所有元素都初始化为 0。
• 对于给定数组中的每个元素 arr[i] ，将 1 添加到b[i+1]并从b[min(i + arr[i], N – 1)+ 1] 中减去。
• 然后，计算数组b[] 的前缀和。
• 最后， arr 更新为arr[i] = arr[i] + b[i] 。

下面是上述方法的实现：

C++

// C++ program to find elements in an array
// after performing range updates.
 
#include 
using namespace std;
 
// Function to perform the update on the given
// input array arr[].
void update(vector& arr,
            vector& d, int n)
{
 
    // A new array of size N+1 is defined
    // and 1's are added in that array
    for (int i = 0; i < n - 1; i++) {
        d[i + 1] += 1;
        d[min(i + arr[i], n - 1) + 1] -= 1;
    }
 
    // Loop to perform the prefix sum
    // on the array d[].
    for (int i = 1; i < n; i++) {
        d[i] = d[i] + d[i - 1];
    }
}
 
// Function to print the final
// array after updation
void print(vector& arr,
           vector& d, int n)
{
 
    // Loop to add the values of d[i]
    // to arr[i]
    for (int i = 0; i < n; i++)
        cout << arr[i] + d[i] << " ";
}
 
// Function to perform modified range sum
void modifiedRangeSum(vector& arr, int n)
{
 
    vector d;
 
    // Loop to add N+1 0's in array d[]
    for (int i = 0; i <= n; i++)
        d.push_back(0);
 
    update(arr, d, n);
    print(arr, d, n);
}
 
// Driver code
int main()
{
    vector arr = { 5, 4, 1, 3, 2 };
    int n = 5;
 
    modifiedRangeSum(arr, n);
 
    return 0;
}

Java

// Java program to find elements in an array
// after performing range updates.
import java.util.*;
 
class GFG{
 
static ArrayList arr = new
        ArrayList(Arrays.asList(5, 4, 1, 3, 2));
static int n = 5;
 
// Function to perform the update on the given
// input array arr[].
static void update(ArrayList d)
{
 
    // A new array of size N+1 is defined
    // and 1's are added in that array
    for (int i = 0; i < n - 1; i++) {
        d.set(i + 1,d.get(i+1)+1);
        int x = Math.min(i + arr.get(i), n - 1)+ 1;
        d.set(x,d.get(x)-1);
    }
 
    // Loop to perform the prefix sum
    // on the array d[].
    for (int i = 1; i < n; i++) {
        d.set(i,d.get(i)+d.get(i - 1));
    }
}
 
// Function to print the final
// array after updation
static void print(ArrayList d)
{
 
    // Loop to add the values of d[i]
    // to arr[i]
    for (int i = 0; i < n; i++)
        System.out.print(arr.get(i) + d.get(i)+ " ");
}
 
// Function to perform modified range sum
static void modifiedRangeSum()
{
 
    ArrayList d = new ArrayList();
 
    // Loop to add N+1 0's in array d[]
    for (int i = 0; i <= n; i++)
        d.add(0);
 
    update(d);
    print(d);
}
 
// Driver code
public static void main(String args[])
{
    modifiedRangeSum();
}
}
 
// This code is contributed by Surendra_Gangwar

Python3

# Python3 program to find elements in an array
# after performing range updates.
arr = []
d = []
 
# Function to perform the update on the given
# input array arr[].
def update( n):
     
    global d
    global arr
 
    # A new array of size N+1 is defined
    # and 1's are added in that array
    for i in range(n - 1):
        d[i + 1] += 1
        d[min(i + arr[i], n - 1) + 1] -= 1
     
    # Loop to perform the prefix sum
    # on the array d[].
    for i in range(n):
        d[i + 1] = d[i + 1] + d[i ]
     
# Function to print the final
# array after updation
def print_( n):
     
    global d
    global arr
 
    # Loop to add the values of d[i]
    # to arr[i]
    for i in range(n):
        x = (arr[i] + d[i] )
        print(x, end = " ")
 
# Function to perform modified range sum
def modifiedRangeSum( n):
 
    global d
    global arr
 
    d = []
     
    # Loop to add N+1 0's in array d[]
    for i in range(n + 1):
        d.append(0)
 
    update( n)
    print_(n)
 
# Driver code
arr = [5, 4, 1, 3, 2]
n = 5
 
modifiedRangeSum( n)
 
# This code is contributed by Arnab Kundu

C#

// C# program to find elements in an array
// after performing range updates.
 
using System;
       
class GFG {
 
// Function to perform the update on the given
// input array arr[].
static void update(int []arr,int [] d, int n){
  
    // A new array of size N+1 is defined
    // and 1's are added in that array
    for (int i = 0; i < n - 1; i++) {
        d[i + 1] += 1;
        d[Math.Min(i + arr[i], n - 1) + 1] -= 1;
    }
  
    // Loop to perform the prefix sum
    // on the array d[].
    for (int i = 1; i < n; i++) {
        d[i] = d[i] + d[i - 1];
    }
}
  
// Function to print the final
// array after updation
static void print(int []arr,int []d, int n)
{
  
    // Loop to add the values of d[i]
    // to arr[i]
    for (int i = 0; i < n; i++)
        Console.Write((arr[i] + d[i])+" ");
}
  
// Function to perform modified range sum
static void modifiedRangeSum(int []arr, int n)
{
    int []d= new int[n+1];
  
    // Loop to add N+1 0's in array d[]
    for (int i = 0; i <= n; i++)
        d[i]=0;
  
    update(arr, d, n);
    print(arr, d, n);
}
 
// Driver code
public static void Main()
  {
    int [] arr = { 5, 4, 1, 3, 2 };
    int n = 5;
  
    modifiedRangeSum(arr, n);
  }
} 
 
// This code is contributed by mohit kumar 29

Javascript

 
// Javascript program to find elements in an array
// after performing range updates.
 
// Function to perform the update on the given
// input array arr[].
function update(arr, d, n)
{
 
    // A new array of size N+1 is defined
    // and 1's are added in that array
    for (var i = 0; i < n - 1; i++) {
        d[i + 1] += 1;
        d[Math.min(i + arr[i], n - 1) + 1] -= 1;
    }
 
    // Loop to perform the prefix sum
    // on the array d[].
    for (var i = 1; i < n; i++) {
        d[i] = d[i] + d[i - 1];
    }
}
 
// Function to print the final
// array after updation
function print(arr, d, n)
{
 
    // Loop to add the values of d[i]
    // to arr[i]
    for (var i = 0; i < n; i++)
        document.write( arr[i] + d[i] + " ");
}
 
// Function to perform modified range sum
function modifiedRangeSum(arr, n)
{
 
    var d = [];
 
    // Loop to add N+1 0's in array d[]
    for (var i = 0; i <= n; i++)
        d.push(0);
 
    update(arr, d, n);
    print(arr, d, n);
}
 
// Driver code
var arr = [5, 4, 1, 3, 2];
var n = 5;
modifiedRangeSum(arr, n);
 
// This code is contributed by importantly.

输出：

5 5 3 6 5

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