给定一个数组arr [] ，该数组arr []由N个正整数和一个整数K组成，整数K表示可以添加到数组元素的最大数量。任务是通过最多添加K来最大化相等元素的最长可能子数组的长度。
例子：
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.
方法：请按照以下步骤解决问题：
- 对数组arr []排序。然后，使用二进制搜索为具有相同元素的最大索引选择一个可能的值。
- 对于每个picked_value，请使用“滑动窗口”技术检查是否有可能使所有元素对于大小为picked_value的子数组相等。
- 最后，打印获得的子数组的最大可能长度。
下面是上述方法的实现：
C++14
// C++14 program for above approach
#include
using namespace std;
// Function to check if a subarray of
// length len consisting of equal elements
// can be obtained or not
bool check(vector pSum, int len, int k,
vector 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 arr, int k)
{
// Sort the array in
// ascending order
sort(arr.begin(), arr.end());
// Make prefix sum array
vector 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 arr = { 1, 1, 1 };
int k = 7;
// Function call
cout << (maxEqualIdx(arr, k));
}
// This code is contributed by mohit kumar 29 Java
// 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
# 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_huntC#
// 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输出：
3
时间复杂度： O(N * log N)
辅助空间： O(N)