给定一个由前N 个自然数排列组成的数组arr[] ,任务是找到使满足条件arr[i] = i ( 1-基于索引),其中1 ≤ i ≤ N。
例子:
Input: arr[] = {4, 5, 1, 2, 3}
Output: 3
Explanation: Rotating the array thrice, the array modifies to {1, 2, 3, 4, 5}. All the array elements satisfy the condition arr[i] = i.
Input: arr[] = {3, 4, 1, 5, 2}
Output: 2
Explanation: Rotating the array twice, the array modifies to {5, 2, 3, 4, 1}. Three array elements satisfy the condition arr[i] = i, which is the maximum possible for the given array.
处理方法:按照以下步骤解决问题:
- 初始化两个整数maxi和ans ,以及两个数组new_arr[]和freq[] 。
- 为每个元素遍历数组arr[] an,计算将其与其正确位置分开的索引数量,即|(arr[i] – i + N) % N| .
- 将每个数组元素的计数存储在新数组new_arr[] 中。
- 将new_arr[]中每个元素的频率计数存储在数组freq[] 中。
- 打印元素 i 最大频率作为所需答案。
下面是上述方法的实现:
C++
// C++ program for the above approach
#include
using namespace std;
// Function to count the number of
// clockwise array rotations required
// to maximize count of array elements
// present at indices same as their value
void find_min_rot(int arr[], int n)
{
// Stores count of indices separating
// elements from its correct position
int new_arr[n + 1];
int maxi = 1, ans = 0;
// Stores frequencies of counts of
// indices separating
int freq[n + 1];
for (int i = 1; i <= n; i++) {
freq[i] = 0;
}
// Count indices separating each
// element from its correct position
for (int i = 1; i <= n; i++) {
new_arr[i] = (arr[i] - i + n) % n;
}
// Update frequencies of counts obtained
for (int i = 1; i <= n; i++) {
freq[new_arr[i]]++;
}
// Find the count with maximum frequency
for (int i = 1; i <= n; i++) {
if (freq[i] > maxi) {
maxi = freq[i];
ans = i;
}
}
// Print the answer
cout << ans << endl;
}
// Driver Code
int main()
{
int N = 5;
int arr[] = { -1, 3, 4, 1, 5, 2 };
// Find minimum number of
// array rotations required
find_min_rot(arr, N);
return 0;
}
Java
// Java program for the above approach
import java.util.*;
class GFG{
// Function to count the number of
// clockwise array rotations required
// to maximize count of array elements
// present at indices same as their value
static void find_min_rot(int arr[], int n)
{
// Stores count of indices separating
// elements from its correct position
int[] new_arr = new int[n + 1];
int maxi = 1, ans = 0;
// Stores frequencies of counts of
// indices separating
int[] freq = new int[n + 1];
for(int i = 1; i <= n; i++)
{
freq[i] = 0;
}
// Count indices separating each
// element from its correct position
for(int i = 1; i <= n; i++)
{
new_arr[i] = (arr[i] - i + n) % n;
}
// Update frequencies of counts obtained
for(int i = 1; i <= n; i++)
{
freq[new_arr[i]]++;
}
// Find the count with maximum frequency
for(int i = 1; i <= n; i++)
{
if (freq[i] > maxi)
{
maxi = freq[i];
ans = i;
}
}
// Print the answer
System.out.print(ans);
}
// Driver Code
public static void main(String[] args)
{
int N = 5;
int[] arr = { -1, 3, 4, 1, 5, 2 };
// Find minimum number of
// array rotations required
find_min_rot(arr, N);
}
}
// This code is contributed by sanjoy_62
Python3
# Python3 program for the above approach
# Function to count the number of
# clockwise array rotations required
# to maximize count of array elements
# present at indices same as their value
def find_min_rot(arr, n):
# Stores count of indices separating
# elements from its correct position
new_arr = [0] * (n + 1)
maxi = 1
ans = 0
# Stores frequencies of counts of
# indices separating
freq = [0] * (n + 1)
for i in range(1, n + 1):
freq[i] = 0
# Count indices separating each
# element from its correct position
for i in range(1, n + 1):
new_arr[i] = (arr[i] - i + n) % n
# Update frequencies of counts obtained
for i in range(1, n + 1):
freq[new_arr[i]] += 1
# Find the count with maximum frequency
for i in range(1, n + 1):
if (freq[i] > maxi):
maxi = freq[i]
ans = i
# Print the answer
print(ans)
# Driver Code
if __name__ == '__main__':
N = 5
arr = [ -1, 3, 4, 1, 5, 2 ]
# Find minimum number of
# array rotations required
find_min_rot(arr, N)
# This code is contributed by jana_sayantan
C#
// C# program for the above approach
using System;
class GFG
{
// Function to count the number of
// clockwise array rotations required
// to maximize count of array elements
// present at indices same as their value
static void find_min_rot(int []arr, int n)
{
// Stores count of indices separating
// elements from its correct position
int[] new_arr = new int[n + 1];
int maxi = 1, ans = 0;
// Stores frequencies of counts of
// indices separating
int[] freq = new int[n + 1];
for(int i = 1; i <= n; i++)
{
freq[i] = 0;
}
// Count indices separating each
// element from its correct position
for(int i = 1; i <= n; i++)
{
new_arr[i] = (arr[i] - i + n) % n;
}
// Update frequencies of counts obtained
for(int i = 1; i <= n; i++)
{
freq[new_arr[i]]++;
}
// Find the count with maximum frequency
for(int i = 1; i <= n; i++)
{
if (freq[i] > maxi)
{
maxi = freq[i];
ans = i;
}
}
// Print the answer
Console.Write(ans);
}
// Driver Code
public static void Main(String[] args)
{
int N = 5;
int[] arr = { -1, 3, 4, 1, 5, 2 };
// Find minimum number of
// array rotations required
find_min_rot(arr, N);
}
}
// This code is contributed by 29AjayKumar
Javascript
输出:
2
时间复杂度: O(N)
辅助空间: O(N)
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