Find the value of K after replacing every index of the array by |ai - K|
Last Updated :
23 Aug, 2023
Given an array, arr[] containing n integers, the task is to find an integer (say K) such that after replacing each and every index of the array by |ai - K| where ( i ? [1, n]), results in a sorted array. If no such integer exists that satisfies the above condition then return -1.
Examples:
Input: arr[ ] = [10, 5, 4, 3, 2, 1]
Output: 8
Explanation: Upon performing the operation |ai-8|, we get [2, 3, 4, 5, 6, 7] which is sorted.
Input: arr[ ] = [1, 2, 3, 4, 5, 6]
Output: 0
Explanation: Since the array is already sorted so the value of K would be 0 in this case.
Approach: The problem can be solved based on the following observation:
Observations:
For the array to be sorted each pair of adjacent elements should be sorted. That means few cases arise if we take care for particular ai and ai+1 and those are as follows:
- Let (ai < ai+1 ), so the following inequality arise:
- If (K ?ai) then upon (ai - K ? ai+1 - K) the elements will be as it is (ai < ai+1).
- If (K ? ai) then upon ( K - ai ? K - ai+1 ) the elements will be as it is (ai > ai+1).
- So, K should be midway between ai and ai+1 that is K should be K ? (ai + ai+1)/2 .
- Similarly for (ai > ai+1) the value of k would be K ? (ai + ai+1)/2
- Finally we will take the minimum of all for which (K < ai) and maximum of all for which (K > ai).
Follow the steps mentioned below to implement the idea:
- Initialize two variables l and r for the two values of k explained above.
- Iterate over the array and check if (ai < ai+1) then store in r the minimum of r so far and the present value of K.
- Else if, (ai > ai+1) stores it in l the maximum of l so far and the present value of K.
- Finally if (l > r) return -1;
- Else, return l
Below is the Implementation of the above approach:
C++
// C++ code for the above approach:
#include <bits/stdc++.h>
using namespace std;
// Function to find K
void findk(int a[], int n)
{
// Initializing the two variables
int l = 0, r = 1e9;
for (int i = 0; i < n - 1; i++) {
// If (a[i] < a[i+1]) then take
// minimum and store in variable
if (a[i] < a[i + 1]) {
r = min(r, (a[i] + a[i + 1]) / 2);
}
// If (a[i]>a[i+1]) then take
// maximum and store in
// separate variable
else if (a[i] > a[i + 1]) {
l = max(l, (a[i] + a[i + 1] + 1) / 2);
}
}
if (l > r) {
cout << "-1";
}
else
cout << l << endl;
}
// Driver function
int main()
{
int arr[] = { 10, 5, 4, 3, 2, 1 };
int n = sizeof(arr) / sizeof(arr[0]);
// Function Call
findk(arr, n);
return 0;
}
// Java Implementation
import java.util.*;
class GFG {
public static void main(String[] args)
{
int[] arr = { 10, 5, 4, 3, 2, 1 };
int n = arr.length;
find(arr, n);
}
public static void find(int[] arr, int n)
{
// Initializing the two variables
int l = 0;
int r = 1000000000;
for (int i = 0; i < n - 1; i++) {
// If (a[i]<a[i+1]) then take minimum and store in variable)
if (arr[i] < arr[i + 1]) {
r = Math.min(r, (arr[i] + arr[i + 1]) / 2);
}
// If (a[i]>a[i+1]) then take maximum and store in separate variable)
else if (arr[i] > arr[i + 1]) {
l = Math.max(l,
(arr[i] + arr[i + 1] + 1) / 2);
}
}
if (l > r) {
System.out.println(-1);
}
else {
System.out.println(l);
}
}
}
# Python code for the above approach
# Function to find K
def findk(a, n):
# Initializing the two variables
l = 0
r = 10**9
for i in range(n - 1):
# If (a[i] < a[i+1]) then take
# minimum and store in variable
if (a[i] < a[i + 1]):
r = min(r, (a[i] + a[i + 1]) // 2)
# If (a[i]>a[i+1]) then take
# maximum and store in
# separate variable
elif (a[i] > a[i + 1]):
l = max(l, (a[i] + a[i + 1] + 1) // 2)
if (l > r):
print("-1")
else:
print(l)
# Driver function
if __name__ == '__main__':
arr = [10, 5, 4, 3, 2, 1]
n = len(arr)
# Function Call
findk(arr, n)
// C# code for the above approach:
using System;
public class GFG {
static public void Main()
{
// Code
int[] arr = { 10, 5, 4, 3, 2, 1 };
int n = arr.Length;
find(arr, n);
}
static void find(int[] arr, int n)
{
// Initializing the two variables
int l = 0;
int r = 1000000000;
for (int i = 0; i < n - 1; i++)
{
// If (a[i]<a[i+1]) then take minimum and store
// in variable)
if (arr[i] < arr[i + 1]) {
r = Math.Min(r, (arr[i] + arr[i + 1]) / 2);
}
// If (a[i]>a[i+1]) then take maximum and store
// in separate variable)
else if (arr[i] > arr[i + 1]) {
l = Math.Max(l,
(arr[i] + arr[i + 1] + 1) / 2);
}
}
if (l > r) {
Console.WriteLine(-1);
}
else {
Console.WriteLine(l);
}
}
}
// This code is contributed by karthik
// JavaScript code for the above approach:
// Function to find K
function findk(a, n) {
// Initializing the two variables
let l = 0, r = 1e9;
for (let i = 0; i < n - 1; i++) {
// If (a[i] < a[i+1]) then take
// minimum and store in variable
if (a[i] < a[i + 1]) {
r = Math.min(r, Math.floor((a[i] + a[i + 1]) / 2));
}
// If (a[i]>a[i+1]) then take
// maximum and store in
// separate variable
else if (a[i] > a[i + 1]) {
l = Math.max(l, Math.ceil((a[i] + a[i + 1]) / 2));
}
}
if (l > r) {
console.log("-1");
}
else {
console.log(l);
}
}
// Driver function
var arr = [10, 5, 4, 3, 2, 1];
var n = arr.length;
// Function Call
findk(arr, n);
// This code is contributed by prasad264
Time Complexity: O(n), iterating the loop for once only.
Auxiliary Space: O(1), no extra space is used.
Another Approach:
- Find the maximum and minimum values in the array.
- Perform binary search to find the value of K such that the absolute difference between adjacent elements of the resulting array is either 0 or 1.
- In each iteration of binary search, calculate the absolute difference between adjacent elements of the resulting array.
- If the absolute difference is greater than 1, set the search range accordingly.
- If the absolute difference is less than or equal to 1, update the answer with the current value of K and set the search range accordingly.
C++
#include <bits/stdc++.h>
using namespace std;
int findK(int arr[], int n) {
int lo = *min_element(arr, arr+n);
int hi = *max_element(arr, arr+n);
int ans = -1;
while (lo <= hi) {
int mid = lo + (hi - lo) / 2;
bool valid = true;
for (int i = 1; i < n; i++) {
if (abs(arr[i] - mid) < abs(arr[i-1] - mid)) {
valid = false;
break;
}
}
if (valid) {
ans = mid;
hi = mid - 1;
} else {
lo = mid + 1;
}
}
return ans;
}
int main() {
int arr[] = {10, 5, 4, 3, 2, 1};
int n = sizeof(arr) / sizeof(arr[0]);
int k = findK(arr, n);
if (k == -1) {
cout << "No such K exists\n";
} else {
cout << k;
}
return 0;
}
import java.util.Arrays;
public class Main {
// Function to find K
public static int findK(int[] arr, int n) {
// Finding minimum and maximum from array
int lo = Arrays.stream(arr).min().getAsInt();
int hi = Arrays.stream(arr).max().getAsInt();
int ans = -1;
// Binary search to find value that absolute difference between adjacent element
// of resulting array is either 0 or 1
while (lo <= hi) {
int mid = lo + (hi - lo) / 2;
boolean valid = true;
// Inner loop calculating the absolute differece between adjacent elements of the resulting
// array.
for (int i = 1; i < n; i++) {
if (Math.abs(arr[i] - mid) < Math.abs(arr[i-1] - mid)) {
valid = false;
break;
}
}
// if difference is less than 1 update value of ans.
if (valid) {
ans = mid;
hi = mid - 1;
} else {
lo = mid + 1;
}
}
return ans;
}
public static void main(String[] args) {
// Input
int[] arr = {10, 5, 4, 3, 2, 1};
int n = arr.length;
int k = findK(arr, n);
if (k == -1) {
System.out.println("No such K exists");
} else {
System.out.println(k);
}
}
}
import sys
def findK(arr, n):
lo = min(arr)
hi = max(arr)
ans = -1
while lo <= hi:
mid = lo + (hi - lo) // 2
valid = True
for i in range(1, n):
if abs(arr[i] - mid) < abs(arr[i-1] - mid):
valid = False
break
if valid:
ans = mid
hi = mid - 1
else:
lo = mid + 1
return ans
arr = [10, 5, 4, 3, 2, 1]
n = len(arr)
k = findK(arr, n)
if k == -1:
print("No such K exists")
else:
print(k)
using System;
using System.Linq;
class Program
{
static int FindK(int[] arr, int n)
{
// Finding min and max of array and store in lo and hi respectively
int lo = arr.Min();
int hi = arr.Max();
int ans = -1;
// Binary serach to find the element such that the difference
// adjacent elements of the resulting array is either 0 or 1
while (lo <= hi)
{
int mid = lo + (hi - lo) / 2;
bool valid = true;
for (int i = 1; i < n; i++)
{
if (Math.Abs(arr[i] - mid) < Math.Abs(arr[i - 1] - mid))
{
valid = false;
break;
}
}
if (valid)
{
ans = mid;
hi = mid - 1;
}
else
{
lo = mid + 1;
}
}
// Returning ans
return ans;
}
// Driver code
static void Main(string[] args)
{
int[] arr = { 10, 5, 4, 3, 2, 1 };
int n = arr.Length;
int k = FindK(arr, n);
if (k == -1)
{
Console.WriteLine("No such K exists");
}
else
{
Console.WriteLine(k);
}
}
}
// Function to find K
function findK(arr, n) {
// Finding minimum and maximum from array
let lo = Math.min(...arr);
let hi = Math.max(...arr);
let ans = -1;
// Binary search to find value that absolute difference between adjacent element
// of resulting array is either 0 or 1
while (lo <= hi) {
let mid = lo + Math.floor((hi - lo) / 2);
let valid = true;
// Inner loop calculating the absolute differece between adjacent elements of the resulting
// array.
for (let i = 1; i < n; i++) {
if (Math.abs(arr[i] - mid) < Math.abs(arr[i-1] - mid)) {
valid = false;
break;
}
}
// if difference is less than 1 update value of ans.
if (valid) {
ans = mid;
hi = mid - 1;
} else {
lo = mid + 1;
}
}
return ans;
}
// Test case
let arr = [10, 5, 4, 3, 2, 1];
let n = arr.length;
let k = findK(arr, n);
if (k === -1) {
console.log("No such K exists");
} else {
console.log(k);
}
Time Complexity: O(n log (hi - lo))
Auxiliary Space: O(1)
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