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Print modified array after performing queries to add (i – L + 1) to each element present in the range [L, R]
  • Last Updated : 22 Apr, 2021

Given an array arr[] consisting of N 0s (1-based indexing) and another array query[], with each row of the form {L, R}, the task for each query (L, R) is to add a value of (i – L + 1) over the range [L, R] and print the array arr[] obtained after performing all the queries.

Examples:

Input: arr[] = {0, 0, 0}, query[][] = {{1, 3}, {2, 3}}
Output: 1 3 5
Explanation: Initially the array is {0, 0, 0}.
Query 1: Range of indices involved: [1, 3]. The value (i – 1 + 1) for each index i in the range is {1, 2, 3}. Adding these values modifies the array to {1, 2, 3}.
Query 2: Range of indices involved: [2, 3]. The value (i – 2 + 1) for each index i in the range is {0, 1, 2}. Adding these values modifies the array to {1, 3, 5}. 
Therefore, the modified array is {1, 3, 5}.

Input: arr[] = {0, 0, 0, 0, 0, 0, 0}, query[][] = {{1, 7}, {3, 6}, {4, 5}}
Output: 1 2 4 7 10 10 7

Naive Approach: The simplest approach to solve the given problem is to traverse the given array over the range [L, R] and add the value (i  –  L  +  1) to each element in the range for every query. After completing all the queries, print the modified array obtained arr[].



Below is the implementation of the above approach:

C++




// C++ program for the above approach
 
#include <bits/stdc++.h>
using namespace std;
 
// Function to perform the given queries
// in the given empty array of size N
void updateQuery(vector<vector<int> > queries,
                 int N)
{
    // Initialize an array a[]
    vector<int> a(N + 1, 0);
 
    // Stores the size of the array
    int n = N + 1;
 
    int q = queries.size();
 
    // Traverse the queries
    for (int i = 0; i < q; i++) {
 
        // Starting index
        int l = queries[i][0];
 
        // Ending index
        int r = queries[i][1];
 
        // Increment each index from L to
        // R in a[] by (j - l + 1)
        for (int j = l; j <= r; j++) {
            a[j] += (j - l + 1);
        }
    }
 
    // Print the modified array
    for (int i = 1; i <= N; i++) {
        cout << a[i] << " ";
    }
}
 
// Driver Code
int main()
{
    int N = 7;
    vector<vector<int> > queries
        = { { 1, 7 }, { 3, 6 }, { 4, 5 } };
 
    // Function Call
    updateQuery(queries, N);
 
    return 0;
}

Java




// Java program for the above approach
import java.util.*;
 
class GFG{
     
// Function to perform the given queries
// in the given empty array of size N
static void updateQuery(int [][]queries, int N)
{
     
    // Initialize an array a[]
    ArrayList<Integer> a = new ArrayList<Integer>();
    for(int i = 0; i < N + 1; i++)
        a.add(0);
 
    // Stores the size of the array
    int q = 3;
 
    // Traverse the queries
    for(int i = 0; i < q; i++)
    {
         
        // Starting index
        int l = queries[i][0];
 
        // Ending index
        int r = queries[i][1];
 
        // Increment each index from L to
        // R in a[] by (j - l + 1)
        for(int j = l; j <= r; j++)
        {
            a.set(j, a.get(j)+(j - l + 1));
        }
    }
 
    // Print the modified array
    for(int i = 1; i < a.size(); i++)
    {
        System.out.print(a.get(i) + " ");
    }
}
 
// Driver code
public static void main(String[] args)
{
    int N = 7;
    int[][] queries =  { { 1, 7 },
                         { 3, 6 },
                         { 4, 5 } };
                                  
    // Function Call
    updateQuery(queries, N);
}
}
 
// This code is contributed by offbeat

Python3




# Python 3 program for the above approach
 
# Function to perform the given queries
# in the given empty array of size N
def updateQuery(queries, N):
   
    # Initialize an array a[]
    a = [0 for i in range(N + 1)]
 
    # Stores the size of the array
    n = N + 1
 
    q = len(queries)
 
    # Traverse the queries
    for i in range(q):
       
        # Starting index
        l = queries[i][0]
 
        # Ending index
        r = queries[i][1]
 
        # Increment each index from L to
        # R in a[] by (j - l + 1)
        for j in range(l, r + 1, 1):
            a[j] += (j - l + 1)
 
    # Print the modified array
    for i in range(1, N + 1, 1):
        print(a[i], end = " ")
 
# Driver Code
if __name__ == '__main__':
    N = 7
    queries =  [[1, 7],[3, 6],[4, 5]]
 
    # Function Call
    updateQuery(queries, N)
 
    # This code is contributed by ipg2016107.

C#




// C# program for the above approach
using System;
using System.Collections.Generic;
 
class GFG{
   
// Function to perform the given queries
// in the given empty array of size N
static void updateQuery(int [,]queries, int N)
{
     
    // Initialize an array a[]
    List<int> a = new List<int>();
    for(int i = 0; i < N + 1; i++)
        a.Add(0);
 
    // Stores the size of the array
    int q = 3;
 
    // Traverse the queries
    for(int i = 0; i < q; i++)
    {
         
        // Starting index
        int l = queries[i, 0];
 
        // Ending index
        int r = queries[i, 1];
 
        // Increment each index from L to
        // R in a[] by (j - l + 1)
        for(int j = l; j <= r; j++)
        {
            a[j] += (j - l + 1);
        }
    }
 
    // Print the modified array
    for(int i = 1; i < a.Count; i++)
    {
        Console.Write(a[i] + " ");
    }
}
 
// Driver Code
public static void Main()
{
    int N = 7;
    int[,] queries = new int[3, 2] { { 1, 7 },
                                     { 3, 6 },
                                     { 4, 5 } };
                                      
    // Function Call
    updateQuery(queries, N);
}
}
 
// This code is contributed by SURENDRA_GANGWAR
Output: 
1 2 4 7 10 10 7

 

Time Complexity: O(N*Q)
Auxiliary Space: O(1)

Efficient Approach: The above approach can be optimized by using the Prefix Sum. Follow the steps below to solve this problem:

  • Initialize an array ans[] with all elements as 0 to stores the number of times the current index is affected by the queries.
  • Initialize an array res[] with all elements as 0 to stores the value to be deleted after the end of a certain query range is reached.
  • Traverse the given array of queries query[] and perform the following steps:
    • Add the 1 to ans[query[i][0]] and subtract 1 from ans[query[i][1] + 1].
    • Subtract (query[i][1] – query[i][0] + 1) from res[query[i][1] + 1].
  • Find the prefix sum of the array ans[].
  • Traverse the array res[] and update each element res[i] as res[i] + res[i – 1] + ans[i].
  • After completing the above steps, print the array res[] as the modified array after performing the given queries.

Below is the implementation of the above approach:

C++




// C++ program for the above approach
 
#include <bits/stdc++.h>
using namespace std;
 
// Function to perform the given queries
// in the given empty array of size N
void updateQuery(vector<vector<int> > queries,
                 int N)
{
    // Stores the resultant array
    // and the difference array
    vector<int> ans(N + 2, 0),
        res(N + 2, 0);
 
    int q = queries.size();
 
    // Traverse the given queries
    for (int i = 0; i < q; i++) {
 
        // Starting index
        int l = queries[i][0];
 
        // Ending index
        int r = queries[i][1];
 
        // Increment l-th index by 1
        ans[l]++;
 
        // Decrease r-th index by 1
        ans[r + 1]--;
 
        // Decrease (r + 1)th index by
        // the length of each query
        res[r + 1] -= (r - l + 1);
    }
 
    // Find the prefix sum of ans[]
    for (int i = 1; i <= N; i++)
        ans[i] += ans[i - 1];
 
    // Find the final array
    for (int i = 1; i <= N; i++)
        res[i] += res[i - 1] + ans[i];
 
    // Printing the modified array
    for (int i = 1; i <= N; i++) {
        cout << res[i] << " ";
    }
    cout << "\n";
}
 
// Driver Code
int main()
{
    int N = 7;
    vector<vector<int> > queries
        = { { 1, 7 }, { 3, 6 }, { 4, 5 } };
 
    updateQuery(queries, N);
 
    return 0;
}
Output: 
1 2 4 7 10 10 7

 

Time Complexity: O(N)
Auxiliary Space: O(N)

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