Given an array arr[] consisting of N positive integers and an integer K, which represents the maximum number that can be added to the array elements. The task is to maximize the length of the longest possible subarray of equal elements by adding at most K.
Examples:
Input: arr[] = {3, 0, 2, 2, 1}, k = 3
Output: 4
Explanation:
Step 1: Adding 2 to arr[1] modifies array to {3, 2, 2, 2, 1}
Step 2: Adding 1 to arr[4] modifies array to {3, 2, 2, 2, 2}
Therefore, answer will be 4 ({arr[1], …, arr[4]}).Input: arr[] = {1, 1, 1}, k = 7
Output: 3
Explanation:
All array elements are already equal. Therefore, the length is 3.
Approach: Follow the steps below to solve the problem:
- Sort the array arr[]. Then, use Binary Search to pick a possible value for the maximum indices having the same element.
- For each picked_value, use the Sliding Window technique to check if it is possible to make all elements equal for any subarray of size picked_value.
- Finally, print the longest possible length of subarray obtained.
Below is the implementation for the above approach:
// C++14 program for above approach #include <bits/stdc++.h> using namespace std;
// Function to check if a subarray of // length len consisting of equal elements // can be obtained or not bool check(vector< int > pSum, int len, int k,
vector< int > a)
{ // Sliding window
int i = 0;
int j = len;
while (j <= a.size())
{
// Last element of the sliding window
// will be having the max size in the
// current window
int maxSize = a[j - 1];
int totalNumbers = maxSize * len;
// The current number of element in all
// indices of the current sliding window
int currNumbers = pSum[j] - pSum[i];
// If the current number of the window,
// added to k exceeds totalNumbers
if (currNumbers + k >= totalNumbers)
{
return true ;
}
else
{
i++;
j++;
}
}
return false ;
} // Function to find the maximum number of // indices having equal elements after // adding at most k numbers int maxEqualIdx(vector< int > arr, int k)
{ // Sort the array in
// ascending order
sort(arr.begin(), arr.end());
// Make prefix sum array
vector< int > prefixSum(arr.size());
prefixSum[1] = arr[0];
for ( int i = 1;
i < prefixSum.size() - 1; ++i)
{
prefixSum[i + 1] = prefixSum[i] +
arr[i];
}
// Initialize variables
int max = arr.size();
int min = 1;
int ans = 1;
while (min <= max)
{
// Update mid
int mid = (max + min) / 2;
// Check if any subarray
// can be obtained of length
// mid having equal elements
if (check(prefixSum, mid, k, arr))
{
ans = mid;
min = mid + 1;
}
else
{
// Decrease max to mid
max = mid - 1;
}
}
return ans;
} // Driver Code int main()
{ vector< int > arr = { 1, 1, 1 };
int k = 7;
// Function call
cout << (maxEqualIdx(arr, k));
} // This code is contributed by mohit kumar 29 |
// Java program for above approach import java.util.*;
class GFG {
// Function to find the maximum number of
// indices having equal elements after
// adding at most k numbers
public static int maxEqualIdx( int [] arr,
int k)
{
// Sort the array in
// ascending order
Arrays.sort(arr);
// Make prefix sum array
int [] prefixSum
= new int [arr.length + 1 ];
prefixSum[ 1 ] = arr[ 0 ];
for ( int i = 1 ; i < prefixSum.length - 1 ;
++i) {
prefixSum[i + 1 ]
= prefixSum[i] + arr[i];
}
// Initialize variables
int max = arr.length;
int min = 1 ;
int ans = 1 ;
while (min <= max) {
// Update mid
int mid = (max + min) / 2 ;
// Check if any subarray
// can be obtained of length
// mid having equal elements
if (check(prefixSum, mid, k, arr)) {
ans = mid;
min = mid + 1 ;
}
else {
// Decrease max to mid
max = mid - 1 ;
}
}
return ans;
}
// Function to check if a subarray of
// length len consisting of equal elements
// can be obtained or not
public static boolean check( int [] pSum,
int len, int k,
int [] a)
{
// Sliding window
int i = 0 ;
int j = len;
while (j <= a.length) {
// Last element of the sliding window
// will be having the max size in the
// current window
int maxSize = a[j - 1 ];
int totalNumbers = maxSize * len;
// The current number of element in all
// indices of the current sliding window
int currNumbers = pSum[j] - pSum[i];
// If the current number of the window,
// added to k exceeds totalNumbers
if (currNumbers + k >= totalNumbers) {
return true ;
}
else {
i++;
j++;
}
}
return false ;
}
// Driver Code
public static void main(String[] args)
{
int [] arr = { 1 , 1 , 1 };
int k = 7 ;
// Function call
System.out.println(maxEqualIdx(arr, k));
}
} |
# Python3 program for above approach # Function to find the maximum number of # indices having equal elements after # adding at most k numbers def maxEqualIdx(arr, k):
# Sort the array in
# ascending order
arr.sort()
# Make prefix sum array
prefixSum = [ 0 ] * ( len (arr) + 1 )
prefixSum[ 1 ] = arr[ 0 ]
for i in range ( 1 , len (prefixSum) - 1 , 1 ):
prefixSum[i + 1 ] = prefixSum[i] + arr[i]
# Initialize variables
max = len (arr)
min = 1
ans = 1
while ( min < = max ):
# Update mid
mid = ( max + min ) / / 2
# Check if any subarray
# can be obtained of length
# mid having equal elements
if (check(prefixSum, mid, k, arr)):
ans = mid
min = mid + 1
else :
# Decrease max to mid
max = mid - 1
return ans
# Function to check if a subarray of # length len consisting of equal elements # can be obtained or not def check(pSum, lenn, k, a):
# Sliding window
i = 0
j = lenn
while (j < = len (a)):
# Last element of the sliding window
# will be having the max size in the
# current window
maxSize = a[j - 1 ]
totalNumbers = maxSize * lenn
# The current number of element in all
# indices of the current sliding window
currNumbers = pSum[j] - pSum[i]
# If the current number of the window,
# added to k exceeds totalNumbers
if (currNumbers + k > = totalNumbers):
return True
else :
i + = 1
j + = 1
return False
# Driver Code arr = [ 1 , 1 , 1 ]
k = 7
# Function call print (maxEqualIdx(arr, k))
# This code is contributed by code_hunt |
// C# program for // the above approach using System;
class GFG{
// Function to find the maximum number of // indices having equal elements after // adding at most k numbers public static int maxEqualIdx( int [] arr,
int k)
{ // Sort the array in
// ascending order
Array.Sort(arr);
// Make prefix sum array
int [] prefixSum = new int [arr.Length + 1];
prefixSum[1] = arr[0];
for ( int i = 1;
i < prefixSum.Length - 1; ++i)
{
prefixSum[i + 1] = prefixSum[i] + arr[i];
}
// Initialize variables
int max = arr.Length;
int min = 1;
int ans = 1;
while (min <= max)
{
// Update mid
int mid = (max + min) / 2;
// Check if any subarray
// can be obtained of length
// mid having equal elements
if (check(prefixSum, mid, k, arr))
{
ans = mid;
min = mid + 1;
}
else
{
// Decrease max to mid
max = mid - 1;
}
}
return ans;
} // Function to check if a subarray of // length len consisting of equal elements // can be obtained or not public static bool check( int [] pSum,
int len, int k,
int [] a)
{ // Sliding window
int i = 0;
int j = len;
while (j <= a.Length)
{
// Last element of the sliding window
// will be having the max size in the
// current window
int maxSize = a[j - 1];
int totalNumbers = maxSize * len;
// The current number of element in all
// indices of the current sliding window
int currNumbers = pSum[j] - pSum[i];
// If the current number of the window,
// added to k exceeds totalNumbers
if (currNumbers + k >= totalNumbers)
{
return true ;
}
else
{
i++;
j++;
}
}
return false ;
} // Driver Code public static void Main(String[] args)
{ int [] arr = {1, 1, 1};
int k = 7;
// Function call
Console.WriteLine(maxEqualIdx(arr, k));
} } // This code is contributed by Rajput-Ji |
<script> // Javascript program to implement // the above approach // Function to find the maximum number of
// indices having equal elements after
// adding at most k numbers
function maxEqualIdx(arr, k)
{
// Sort the array in
// ascending order
arr.sort();
// Make prefix sum array
let prefixSum
= new Array(arr.length + 1).fill(0);
prefixSum[1] = arr[0];
for (let i = 1; i < prefixSum.length - 1;
++i) {
prefixSum[i + 1]
= prefixSum[i] + arr[i];
}
// Initialize variables
let max = arr.length;
let min = 1;
let ans = 1;
while (min <= max) {
// Update mid
let mid = Math.floor((max + min) / 2);
// Check if any subarray
// can be obtained of length
// mid having equal elements
if (check(prefixSum, mid, k, arr)) {
ans = mid;
min = mid + 1;
}
else {
// Decrease max to mid
max = mid - 1;
}
}
return ans;
}
// Function to check if a subarray of
// length len consisting of equal elements
// can be obtained or not
function check(pSum, len, k, a)
{
// Sliding window
let i = 0;
let j = len;
while (j <= a.length) {
// Last element of the sliding window
// will be having the max size in the
// current window
let maxSize = a[j - 1];
let totalNumbers = maxSize * len;
// The current number of element in all
// indices of the current sliding window
let currNumbers = pSum[j] - pSum[i];
// If the current number of the window,
// added to k exceeds totalNumbers
if (currNumbers + k >= totalNumbers) {
return true ;
}
else {
i++;
j++;
}
}
return false ;
}
// Driver Code let arr = [ 1, 1, 1 ];
let k = 7;
// Function call
document.write(maxEqualIdx(arr, k));
</script> |
3
Time Complexity: O(N * log N)
Auxiliary Space: O(N)