Find final Array after subtracting maximum from all elements K times
Last Updated :
15 Jun, 2022
Given an array arr[] of N integers and an integer K, the task is to do the below operations with this array K times. Operations are as follows:
- Find the maximum element (say M) from the array.
- Now replace every element with M-arri. where 1 ≤ i ≤ N.
Examples:
Input: N = 6, K = 3, arr[] = {5, 38, 4, 96, 103, 41}
Output: 98 65 99 7 0 62
Explanation:
1st operation => Maximum array element is 103. Subtract 103 from all elements. arr[] = {98, 65, 99, 7, 0, 62}.
2nd operation => Maximum array element is 99. Subtract 99 from all elements. arr[] = {1, 34, 0, 92, 99, 37}
3rd operation => Maximum array element is 99. Subtract 99 from all elements. arr[] = {98, 65, 99, 7, 0, 62}.
This will be the last state.
Input: N = 5, K = 1, arr[] = {8, 4, 3, 10, 15}
Output: 7 11 12 5 0
Naive Approach: Simple approach is to perform the above operations K times. Every time find the maximum element in the array and then update all the elements with the difference from the maximum.
Time complexity: O(N*K).
Auxiliary Space: O(1).
Efficient Approach: The efficient solution to this problem is based on the below observation:
If K is odd then the final answer will be equivalent to the answer after applying the operation only one time and if K is even then the final array will be equivalent to the array after applying operations only two time.
This can be proved by as per the following:
Say maximum = M, and initial array is arr[].
After the operations are performed 1st time:
=> The maximum element of arr[] becomes 0.
=> All other elements are M - arr[i].
=> Say the new array is a1[]. So, a1[i] = M - arr[i].
After the operation are performed 2nd time:
=> Say maximum of a1[] is M1.
=> The maximum element of a1[] becomes 0.
=> All other elements are M1 - a1[i].
=> Say the new array is a2[]. So, a2[i] = M1 - a1[i].
=> Maximum of a2[] will be M1, because the 0 present in a1[] changes to M1 and all other elements are less than M1.
After the operation are performed 3rd time:
=> The maximum is M1.
=> The maximum changes to 0.
=> All other elements are M1 - a2[i] = M1 - (M1 - a1[i]) = a1[i].
=> So the new array is same as a1[].
After the operation are performed 4th time:
=> Say maximum of the array is M1.
=> The maximum element changes to be 0.
=> All other elements are M1 - a1[i].
=> So the new array is the same as a2[]
From the above it is clear that the alternate states of the array are same.
Follow the below steps to implement the observation:
- Take one variable max and store the maximum element of arr.
- If K is odd
- Traverse the array and subtract every element from the maximum element.
- Else
- Traverse the arr and subtract every element from the maximum element and
store the maximum element in max1 variable. - Again traverse the arr and subtract every element from the maximum element.
- Print the final array.
Below is the implementation of the above approach:
C++
// C++ code for the approach
#include <bits/stdc++.h>
using namespace std;
// Function for converting the array
void find(int n, int k, int arr[])
{
// Find the maximum element in array
int max = INT_MIN;
for (int i = 0; i < n; i++) {
if (arr[i] > max) {
max = arr[i];
}
}
// If k is odd
if (k % 2 != 0) {
for (int i = 0; i < n; i++) {
cout << max - arr[i] << " ";
}
}
// If k is even
else {
// Subtract the max from every
// element of array and store
// the next maximum element in max1
int max1 = INT_MIN;
for (int i = 0; i < n; i++) {
arr[i] = max - arr[i];
if (arr[i] > max1) {
max1 = arr[i];
}
}
// Print the output
for (int i = 0; i < n; i++) {
cout << max1 - arr[i] << " ";
}
}
}
// Driver code
int main()
{
int N = 6, K = 3;
int arr[] = { 5, 38, 4, 96, 103, 41 };
// Function call
find(N, K, arr);
return 0;
}
// C code for the approach
#include <stdio.h>
#include<limits.h>
// Function for converting the array
void find(int n, int k, int arr[])
{
// Find the maximum element in array
int max = INT_MIN;
for (int i = 0; i < n; i++) {
if (arr[i] > max) {
max = arr[i];
}
}
// If k is odd
if (k % 2 != 0) {
for (int i = 0; i < n; i++) {
printf("%d ", max - arr[i]);
}
}
// If k is even
else {
// Subtract the max from every
// element of array and store
// the next maximum element in max1
int max1 = INT_MIN;
for (int i = 0; i < n; i++) {
arr[i] = max - arr[i];
if (arr[i] > max1) {
max1 = arr[i];
}
}
// Print the output
for (int i = 0; i < n; i++) {
printf("%d ", max1 - arr[i]);
}
}
}
// Driver code
void main()
{
int N = 6, K = 3;
int arr[] = { 5, 38, 4, 96, 103, 41 };
// Function call
find(N, K, arr);
}
// This code is contributed by ashishsingh13122000.
// Java code for the approach
import java.io.*;
class GFG
{
// Function for converting the array
public static void find(int n, int k, int arr[])
{
// Find the maximum element in array
int max = Integer.MIN_VALUE;
for (int i = 0; i < n; i++) {
if (arr[i] > max) {
max = arr[i];
}
}
// If k is odd
if (k % 2 != 0) {
for (int i = 0; i < n; i++) {
System.out.print((max - arr[i]) + " ");
}
}
// If k is even
else {
// Subtract the max from every
// element of array and store
// the next maximum element in max1
int max1 = Integer.MIN_VALUE;
for (int i = 0; i < n; i++) {
arr[i] = max - arr[i];
if (arr[i] > max1) {
max1 = arr[i];
}
}
// Print the output
for (int i = 0; i < n; i++) {
System.out.print((max1 - arr[i]) + " ");
}
}
}
// Driver Code
public static void main(String[] args)
{
int N = 6, K = 3;
int arr[] = { 5, 38, 4, 96, 103, 41 };
// Function call
find(N, K, arr);
}
}
// This code is contributed by Rohit Pradhan
# Python code
# Function for converting the array
def find(n, k, arr):
# Find the maximum element in array
max = (-2147483647 - 1)
for i in range(0, n):
if (arr[i] > max):
max = arr[i]
# If k is odd
if (k % 2 != 0):
for i in range (0, n):
print( max - arr[i], end = " ")
# If k is even
else:
max1 = INT_MIN
for i in range(0,n):
arr[i] = max - arr[i]
if (arr[i] > max1):
max1 = arr[i]
# Print the output
for i in range(0,n):
print(max1 - arr[i],end=" ")
# Driver code
N = 6
K = 3
arr = [5, 38, 4, 96, 103, 41]
# Function call
find(N,K,arr)
# This code is contributed by ashishsingh13122000.
// C# code for the approach
using System;
class GFG
{
// Function for converting the array
static void find(int n, int k, int []arr)
{
// Find the maximum element in array
int max = Int32.MinValue;
for (int i = 0; i < n; i++) {
if (arr[i] > max) {
max = arr[i];
}
}
// If k is odd
if (k % 2 != 0) {
for (int i = 0; i < n; i++) {
Console.Write((max - arr[i]) + " ");
}
}
// If k is even
else {
// Subtract the max from every
// element of array and store
// the next maximum element in max1
int max1 = Int32.MinValue;
for (int i = 0; i < n; i++) {
arr[i] = max - arr[i];
if (arr[i] > max1) {
max1 = arr[i];
}
}
// Print the output
for (int i = 0; i < n; i++) {
Console.Write((max1 - arr[i]) + " ");
}
}
}
// Driver Code
public static void Main()
{
int N = 6, K = 3;
int []arr = { 5, 38, 4, 96, 103, 41 };
// Function call
find(N, K, arr);
}
}
// This code is contributed by Samim Hossain Mondal.
<script>
// JavaScript code for the approach
const INT_MIN = -2147483648;
// Function for converting the array
const find = (n, k, arr) => {
// Find the maximum element in array
let max = INT_MIN;
for (let i = 0; i < n; i++) {
if (arr[i] > max) {
max = arr[i];
}
}
// If k is odd
if (k % 2 != 0) {
for (let i = 0; i < n; i++) {
document.write(`${max - arr[i]} `);
}
}
// If k is even
else {
// Subtract the max from every
// element of array and store
// the next maximum element in max1
let max1 = INT_MIN;
for (let i = 0; i < n; i++) {
arr[i] = max - arr[i];
if (arr[i] > max1) {
max1 = arr[i];
}
}
// Print the output
for (let i = 0; i < n; i++) {
document.write(`${max1 - arr[i]} `);
}
}
}
// Driver code
let N = 6, K = 3;
let arr = [5, 38, 4, 96, 103, 41];
// Function call
find(N, K, arr);
// This code is contributed by rakeshsahni
</script>
Time Complexity: O(N)
Auxiliary Space: O(1)
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