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Maximum elements that can be made equal with k updates in C++
Given the task is to find the maximum number of elements that can be made equal in a given array after incrementing its elements by at-most k times.
Let’s now understand what we have to do using an example −
Input
a[] = {1, 3, 8}, k = 4Output
2
Explanation
Here we can obtain two fours by incrementing 1 three times and incrementing 3 four times, that makes a[] = {4, 4, 8}
Input
arr = {2, 5, 9}, k = 2Output
0
Approach used in the below program as follows
In main() function initialize int a[], size and k to store the array elements, size of array and the maximum updates possible respectively.
In max() function, first sort the array in ascending order and then declare two more arrays int p[size + 1] and m[size + 1] that will store prefix sum and maximum value respectively.
Loop from i = 0 till i<= size and initialize p[i] = 0 and m[i] = 0.
Outside the loop put m[0] = arr[0] and p[0] = arr[0].
Loop from i = 1 till i<size and put p[i] = p[i - 1] + arr[i] to calculate prefix sum and put m[i] = max(m[i - 1], arr[i]) to calculate maximum value up to that position.
After the loop, initialize int Lt = 1, Rt = size, result to store left value, right value and final answer respectively. After that initiate binary search.
Initiate while loop with condition (Lt < Rt). Inside the loop put int mid = (Lt + Rt) / 2. Check if (EleCal(p, m, mid - 1, k, size)). If so, then put result = mid and Lt = mid +1.
Else simply put Rt = mid -1.
Outside the loop print result.
In function bool EleCal() initiate for loop with condition for (int i = 0, j = x; j <= size; j++, i++)
Inside the loop check if (x * m[j] - (p[j] - p[i]) <= k). If so, then return true. Outside the loop return false.
Example
#include <bits/stdc++.h>
using namespace std;
bool EleCal(int p[], int m[], int x, int k, int size){
for (int i = 0, j = x; j <= size; j++, i++){
if (x * m[j] - (p[j] - p[i]) <= k)
return true;
}
return false;
}
void Max(int arr[], int size, int k){
// sorting array in ascending order
sort(arr, arr + size);
int p[size + 1];
//array for maximum value
int m[size + 1];
// Initialize prefix array and maximum array
for (int i = 0; i <= size; ++i){
p[i] = 0;
m[i] = 0;
}
m[0] = arr[0];
p[0] = arr[0];
for (int i = 1; i < size; i++){
// Calculating prefix sum of the array
p[i] = p[i - 1] + arr[i];
// Calculating max value upto that position
// in the array
m[i] = max(m[i - 1], arr[i]);
}
// Binary seach
int Lt = 1, Rt = size, result;
while (Lt < Rt){
int mid = (Lt + Rt) / 2;
if (EleCal(p, m, mid - 1, k, size)){
result = mid;
Lt = mid + 1;
}
else
Rt = mid - 1;
}
//answer
cout<<result;
}
// main function
int main(){
int a[] = { 1, 3, 8 };
int size = sizeof(a) / sizeof(a[0]);
int k = 4;
Max(a, size, k);
return 0;
}Output
2
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