Given an array arr[] of size N containing distinct numbers from 1 to N in any order, the task is to perform a modified range sum in this array according to the following rules.
For each index ‘i‘ in array arr:
- The starting index of the range ‘L‘ is selected as i + 1
- The ending index of the range ‘R‘ is selected as:
- min(arr[i], N-1); if arr[i] > i
- max(i+1, arr[i]); if arr[i] < i+1
- For updation, the values in range arr[L] to arr[R] is incremented by 1.
- The range is found out using the input array and not the updated array
Examples:
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}.
Naive Approach: The naive approach is to run a loop for each element and increase all the values from arr[i+1] to arr[min(i+arr[i], N-1)] by 1. The time complexity of this approach is O(N2).
Efficient Approach: This problem can be solved in O(N) by using an extra space of O(N). The idea is to use the concept of prefix sum array. The following steps are followed to compute the answer:
- An array b[] of size N + 1 is declared and all the elements are initialized with 0.
- For each element arr[i] in the given array, 1 is added to b[i+1] and subtracted from b[min(i + arr[i], N – 1)+ 1].
- Then, prefix sum of the array b[] is calculated.
- Finally, arr is updated as arr[i] = arr[i] + b[i].
Below is the implementation of the above approach:
// C++ program to find elements in an array // after performing range updates. #include <bits/stdc++.h> using namespace std;
// Function to perform the update on the given // input array arr[]. void update(vector<int>& arr,
vector<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[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<int>& arr,
vector<int>& 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<int>& arr, int n)
{ vector<int> 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<int> arr = { 5, 4, 1, 3, 2 };
int n = 5;
modifiedRangeSum(arr, n);
return 0;
} |
// Java program to find elements in an array // after performing range updates. import java.util.*;
class GFG{
static ArrayList<Integer> arr = new ArrayList<Integer>(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<Integer> 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<Integer> 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<Integer> d = new ArrayList<Integer>();
// 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 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# 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 |
<script> // 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. </script> |
5 5 3 6 5
Time Complexity: O(n)
Auxiliary Space: O(n), where n is the size of the given vector.