Given an array arr[ ] of size N and an integer p, the task is to find maximum possible value of rightmost element of array arr[ ] performing at most k operations.In one operation decrease arr[i] by p and increase arr[i+1] by p.
Examples:
Input: N = 4, p = 2, k = 5, arr = {3, 8, 1, 4}
Output: 8
Explanation:
following operation can be performed to achieve max valued rightmost element:
1. decrease arr[2] by 2 and increase arr[3] by 2, arr = {3, 6, 3, 4}
2. decrease arr[2] by 2 and increase arr[3] by 2, arr = {3, 4, 5, 4}
3. decrease arr[2] by 2 and increase arr[3] by 2, arr = {3, 2, 5, 4}
4. decrease arr[3] by 2 and increase arr[4] by 2, arr = {3, 2, 3, 6}
5. decrease arr[3] by 2 and increase arr[4] by 2, arr = {3, 2, 1, 8}Hence, maximum possible value of rightmost element will be 8.
Input: N = 5, p = 1, k = 2, arr = {1, 2, 3, 4, 5}
Output: 7
Naive Approach: This problem can solved naively using greedy approach. follow the steps below to solve the problem:
- Run a while loop, until is k is greater than 0.
- Run a for loop inside while loop, for i in range [N-2, 0].
- Check if arr[i] is greater than 1, increase arr[i+1] by p and decrease arr[i] by p and break for loop.
- Decrease the value of k by 1.
- Print arr[N-1].
Below is the implementation of the above approach:
C++
// C++ program for above approach#include <iostream>using namespace std;// Function to calculate maximum value of // Rightmost elementvoid maxRightmostElement(int N, int k, int p, int arr[]){ // Calculating maximum value of // Rightmost element while(k) { for(int i=N-2;i>=0;i--) { // Checking if arr[i] is operationable if(arr[i]>= p) { // Performing operation of i-th element arr[i]=arr[i]-p; arr[i+1]=arr[i+1]+p; break; } } // Decreasing the value of k by 1 k--; } // Printing rightmost element cout<<arr[N-1]<<endl;}// Driver Codeint main() { // Given Input int N = 4, p = 2, k = 5, arr[] = {3, 8, 1, 4}; // Function Call maxRightmostElement(N, k, p, arr); return 0;} |
Java
// Java program for the above approachimport java.io.*;class GFG { // Function to calculate maximum value of // Rightmost element static int maxRightmostElement(int N, int k, int p, int arr[]) { // Calculating maximum value of // Rightmost element while (k > 0) { for (int i = N - 2; i >= 0; i--) { // Checking if arr[i] is operationable if (arr[i] >= p) { // Performing operation of i-th element arr[i] = arr[i] - p; arr[i + 1] = arr[i + 1] + p; break; } } // Decreasing the value of k by 1 k--; } // returning the rightmost element return arr[N - 1]; } // Driver Code public static void main(String[] args) { // Given Input int N = 4, k = 5, p = 2; int arr[] = { 3, 8, 1, 4 }; // Function Call System.out.println( maxRightmostElement(N, k, p, arr)); }}// This code is contributed by dwivediyash |
Python3
# Python program for the above approach# Function to calculate maximum value of # Rightmost elementdef maxRightmostElement(N, k, p, arr): # Calculating maximum value of # Rightmost element while(k) : for i in range(N - 2, -1, -1) : # Checking if arr[i] is operationable if(arr[i] >= p) : # Performing operation of i-th element arr[i] = arr[i] - p arr[i + 1] = arr[i + 1] + p break # Decreasing the value of k by 1 k = k - 1 # Printing rightmost element print(arr[N-1])# Driver Code# Given InputN = 4p = 2k = 5arr = [3, 8, 1, 4] # Function CallmaxRightmostElement(N, k, p, arr)# This code is contributed by sanjoy_62. |
C#
using System;public class GFG { // Function to calculate maximum value of // Rightmost element static int maxRightmostElement(int N, int k, int p, int []arr) { // Calculating maximum value of // Rightmost element while (k > 0) { for (int i = N - 2; i >= 0; i--) { // Checking if arr[i] is operationable if (arr[i] >= p) { // Performing operation of i-th element arr[i] = arr[i] - p; arr[i + 1] = arr[i + 1] + p; break; } } // Decreasing the value of k by 1 k--; } // returning the rightmost element return arr[N - 1]; } // Driver Code static public void Main() { // Given Input int N = 4, k = 5, p = 2; int[] arr = { 3, 8, 1, 4 }; // Function Call Console.WriteLine( maxRightmostElement(N, k, p, arr)); }}// This code is contributed by maddler. |
Javascript
<script> // JavaScript Program to implement // the above approach // Function to calculate maximum value of // Rightmost element function maxRightmostElement(N, k, p, arr) { // Calculating maximum value of // Rightmost element while (k) { for (let i = N - 2; i >= 0; i--) { // Checking if arr[i] is operationable if (arr[i] >= p) { // Performing operation of i-th element arr[i] = arr[i] - p; arr[i + 1] = arr[i + 1] + p; break; } } // Decreasing the value of k by 1 k--; } // Printing rightmost element document.write(arr[N - 1] + "<br>"); } // Driver Code // Given Input let N = 4, p = 2, k = 5, arr = [3, 8, 1, 4]; // Function Call maxRightmostElement(N, k, p, arr);// This code is contributed by Potta Lokesh </script> |
Output
8
Time Complexity: O(N*k)
Auxiliary Space: O(1)
Efficient Approach: The above approach can be optimized by observing the fact that, i-th element of array arr[ ] can contribute 2 to the element arr[N-1] in N-1-i operations. follow the steps below to solve the problem:
- Iterate a loop, for i in range [N-2, 0].
- Initialize ans as arr[N-1] to store maximum value of rightmost element.
- Find minimum contribution of i-th element say d as, d = min(a[i]/2, k/(N-1-i)).
- Decrease k by d*(N-1-i) and increase ans by d*2.
- Print ans, it will be final step.
C++
// C++ program for above approach#include <iostream>using namespace std;// Function to calculate maximum value of // Rightmost elementvoid maxRightmostElement(int N, int k, int arr[]){ // Initializing ans to store // Maximum valued rightmost element int ans = arr[N-1]; // Calculating maximum value of // Rightmost element for(int i=N-2; i>=0; i--) { int d = min(arr[i]/2, k/(N-1-i)); k-=d*(N-1-i); ans+=d*2; } // Printing rightmost element cout<<ans<<endl;}// Driver Codeint main() { // Given Input int N = 4, k = 5, arr[] = {3, 8, 1, 4}; // Function Call maxRightmostElement(N, k, arr); return 0;} |
Java
// Java program for the above approachimport java.io.*;class GFG { // Function to calculate maximum value of // Rightmost element static int maxRightmostElement(int N, int k, int p, int arr[]) { // Initializing ans to store // Maximum valued rightmost element int ans = arr[N - 1]; // Calculating maximum value of // Rightmost element for (int i = N - 2; i >= 0; i--) { int d = Math.min(arr[i] / p, k / (N - 1 - i)); k -= d * (N - 1 - i); ans += d * p; } // returning rightmost element return ans; } // Driver Code public static void main(String[] args) { // Given Input int N = 4, k = 5, p = 2; int arr[] = { 3, 8, 1, 4 }; // Function Call System.out.println(maxRightmostElement(N, k, p, arr)); }}// This code is contributed by dwivediyash |
Python3
# Python 3 program for above approach# Function to calculate maximum value of # Rightmost elementdef maxRightmostElement(N, k, arr): # Initializing ans to store # Maximum valued rightmost element ans = arr[N-1] # Calculating maximum value of # Rightmost element i = N - 2 while(i >= 0): d = min(arr[i]//2, k//(N-1-i)) k -= d * (N - 1 - i) ans+=d*2 # Printing rightmost element i -= 1 print(ans,end = " ") # Driver Codeif __name__ == '__main__': # Given Input N = 4 k = 5 arr = [3, 8, 1, 4] # Function Call maxRightmostElement(N, k, arr) # This code is contributed by SURENDRA_GANGWAR. |
C#
// C# program for the above approachusing System;public class GFG { // Function to calculate maximum value of // Rightmost element static int maxRightmostElement(int N, int k, int p, int []arr) { // Initializing ans to store // Maximum valued rightmost element int ans = arr[N - 1]; // Calculating maximum value of // Rightmost element for (int i = N - 2; i >= 0; i--) { int d = Math.Min(arr[i] / p, k / (N - 1 - i)); k -= d * (N - 1 - i); ans += d * p; } // returning rightmost element return ans; } // Driver Code public static void Main(string[] args) { // Given Input int N = 4, k = 5, p = 2; int []arr = { 3, 8, 1, 4 }; // Function Call Console.WriteLine(maxRightmostElement(N, k, p, arr)); }}// This code is contributed by AnkThon |
Javascript
<script>// JavaScript program for the above approach // Function to calculate maximum value of // Rightmost element function maxRightmostElement( N, k, p, arr) { // Initializing ans to store // Maximum valued rightmost element var ans = arr[N - 1]; // Calculating maximum value of // Rightmost element for (var i = N - 2; i >= 0; i--) { var d = Math.min(arr[i] / p, k / (N - 1 - i)); k -= d * (N - 1 - i); ans += d * p; } // returning rightmost element return ans; } // Driver Code // Given Input var N = 4, k = 3.5, p = 2; var arr = [ 3, 8, 1, 4 ]; // Function Call document.write(maxRightmostElement(N, k, p, arr)); // This code is contributed by shivanisinghss2110</script> |
Output
8
Time Complexity: O(N)
Auxiliary Space: O(1)
Feeling lost in the world of random DSA topics, wasting time without progress? It’s time for a change! Join our DSA course, where we’ll guide you on an exciting journey to master DSA efficiently and on schedule.
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
