给定一个由N 个整数组成的数组arr[] ,任务是最大化索引的数量,使元素大于其左侧的元素,即在重新排列数组后arr[i+1] > arr[i] 。
例子:
Input: arr[] = {200, 100, 100, 200}
Output: 2
Explanation:
By arranging the array in following way we have: arr[] = {100, 200, 100, 200}
The possible indices are 0 and 2 such that:
arr[1] > arr[0] (200 > 100)
arr[3] > arr[2] (200 > 100)
Input: arr[] = {1, 8, 5, 9, 8, 8, 7, 7, 5, 7, 7}
Output: 7
Explanation:
By arranging the array in following way we have: arr[] = {1, 5, 7, 8, 9, 5, 7, 8, 7, 8, 4}
The possible indices are 0, 1, 2, 3, 5, 6 and 7 such that:
arr[1] > arr[0] (5 > 1)
arr[2] > arr[1] (7 > 5)
arr[3] > arr[2] (8 > 7)
arr[4] > arr[3] (9 > 8)
arr[6] > arr[5] (7 > 5)
arr[7] > arr[6] (8 > 7)
arr[8] > arr[7] (8 > 7)
方法:这个问题可以使用贪心方法来解决。以下是步骤:
- 要获得索引的最大数量(比如i ),使得arr[i+1] > arr[i] ,排列arr[]的元素,使得所有唯一元素的集合首先出现,然后出现下一组唯一元素在第一组之后,直到所有元素都排列好。
例如:
Let arr[] = {1, 8, 5, 9, 8, 8, 7, 7, 5, 7, 7}
1st Set = {1, 5, 7, 8, 9}
2nd Set = {5, 7, 8}
3rd Set = {7, 8}
4th Set = {4}
Now the new array will be:
arr[] = {1, 5, 7, 8, 9, 5, 7, 8, 7, 8, 4}
- 经过上述排列后,具有较高值的元素将不会成为给定条件的一部分,因为它后面跟着一个小于其自身的数字。
- 因此,满足给定条件的对的总数可以由下式给出:
total_pairs = (number_of_elements – highest_frequency_of_a_number)
下面是上述方法的实现:
C++
// C++ program of the above approach
#include
using namespace std;
// Function to find the maximum pairs
// such that arr[i+1] > arr[i]
void countPairs(int arr[], int N)
{
// To store the frequency of the
// element in arr[]
unordered_map M;
// Store the frequency in map M
for (int i = 0; i < N; i++) {
M[arr[i]]++;
}
int maxFreq = 0;
// To find the maximum frequency
// store in map M
for (auto& it : M) {
maxFreq = max(maxFreq,
it.second);
}
// Print the maximum number of
// possible pairs
cout << N - maxFreq << endl;
}
// Driver Code
int main()
{
int arr[] = { 1, 8, 5, 9, 8, 8, 7,
7, 5, 7, 7 };
int N = sizeof(arr) / sizeof(arr[0]);
countPairs(arr, N);
return 0;
}
Java
// Java program of the above approach
import java.util.*;
class GFG{
// Function to find the maximum pairs
// such that arr[i+1] > arr[i]
static void countPairs(int arr[], int N)
{
// To store the frequency of the
// element in arr[]
HashMap mp = new HashMap();
// Store the frequency in map M
for (int i = 0; i < N; i++) {
if(mp.containsKey(arr[i])){
mp.put(arr[i], mp.get(arr[i])+1);
}else{
mp.put(arr[i], 1);
}
}
int maxFreq = 0;
// To find the maximum frequency
// store in map M
for (Map.Entry it : mp.entrySet()) {
maxFreq = Math.max(maxFreq,
it.getValue());
}
// Print the maximum number of
// possible pairs
System.out.print(N - maxFreq +"\n");
}
// Driver Code
public static void main(String[] args)
{
int arr[] = { 1, 8, 5, 9, 8, 8, 7,
7, 5, 7, 7 };
int N = arr.length;
countPairs(arr, N);
}
}
// This code is contributed by Rajput-Ji
Python3
# Python3 implementation of the above approach
# Function to find the maximum pairs
# such that arr[i + 1] > arr[i]
def countPairs(arr, N) :
# To store the frequency of the
# element in arr[]
M = dict.fromkeys(arr, 0);
# Store the frequency in map M
for i in range(N) :
M[arr[i]] += 1;
maxFreq = 0;
# To find the maximum frequency
# store in map M
for it in M.values() :
maxFreq = max(maxFreq,it);
# Print the maximum number of
# possible pairs
print(N - maxFreq);
# Driver Code
if __name__ == "__main__" :
arr = [ 1, 8, 5, 9, 8, 8, 7, 7, 5, 7, 7 ];
N = len(arr);
countPairs(arr, N);
# This code is contributed by AnkitRai01
C#
// C# program of the above approach
using System;
using System.Collections.Generic;
class GFG{
// Function to find the maximum pairs
// such that arr[i+1] > arr[i]
static void countPairs(int []arr, int N)
{
// To store the frequency of the
// element in []arr
Dictionary mp = new Dictionary();
// Store the frequency in map M
for (int i = 0; i < N; i++) {
if(mp.ContainsKey(arr[i])){
mp[arr[i]] = mp[arr[i]]+1;
}else{
mp.Add(arr[i], 1);
}
}
int maxFreq = 0;
// To find the maximum frequency
// store in map M
foreach (KeyValuePair it in mp) {
maxFreq = Math.Max(maxFreq,
it.Value);
}
// Print the maximum number of
// possible pairs
Console.Write(N - maxFreq +"\n");
}
// Driver Code
public static void Main(String[] args)
{
int []arr = { 1, 8, 5, 9, 8, 8, 7,
7, 5, 7, 7 };
int N = arr.Length;
countPairs(arr, N);
}
}
// This code is contributed by Rajput-Ji
Javascript
7
时间复杂度: O(N),其中 N 是数组中元素的数量。
辅助空间: O(N),其中 N 是数组中元素的数量。
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