森林里有一些彩色的兔子。给定的大小为N的阵列ARR []，使得具有ARR [i]表示兔子的数目相同的颜色的第i个兔子，任务是找到兔子，可能是在森林的最小数量。
例子：
Input: arr[] = {2, 2, 0}
Output: 4
Explanation: Considering the 1st and the 2nd rabbits to be of same color, eg. Blue, there should be 3 blue-colored rabbits. The third rabbit is the only rabbit of that color. Therefore, the minimum number of rabbits that could be present in the forest are = 3 + 1 = 4.
Input: arr[] = {10, 10, 10}
Output: 11
Explanation: Considering all the rabbits to be of the same color, the minimum number of rabbits present in forest are 10 + 1 = 11.
方法：解决这个问题的方法是找出颜色相同的兔子的组数和每组兔子的数量。以下是步骤：
- 初始化一个变量count来存储每组兔子的数量。
- 初始化映射并遍历数组，其键为arr[i] ，值为arr[i]在给定数组中的出现次数。
- 现在，如果y只兔子回答x ，则：
- 如果(y%(x + 1))为0 ，则必须有(y / (x + 1))组(x + 1)只兔子。
- 如果(y % (x + 1))非零，则必须有(y / (x + 1)) + 1组(x + 1)兔子。
- 将组数和每组兔子数的乘积添加到变量count 中。
- 经过上述步骤后， count的值给出了森林中兔子的最小数量。
下面是上述方法的实现：
C++
// C++ program for the above approach
#include
using namespace std;
// Function to find the minimun
// number of rabbits in the forest
int minNumberOfRabbits(int answers[], int N)
{
// Initialize map
map Map;
// Traverse array and map arr[i]
// to the number of occurences
for (int a = 0; a < N; a++) {
Map[answers[a]]++;
}
// Intialize count as 0;
int count = 0;
// Find the number groups and
// no. of rabbits in each group
for (auto a : Map) {
int x = a.first;
int y = a.second;
// Find number of groups and
// multiply them with number
// of rabbits in each group
if (y % (x + 1) == 0)
count = count + (y / (x + 1)) * (x + 1);
else
count = count + ((y / (x + 1)) + 1) * (x + 1);
}
// count gives minimum number
// of rabbits in the forest
return count;
}
// Driver code
int main()
{
int arr[] = { 2, 2, 0 };
int N = sizeof(arr) / sizeof(arr[0]);
// Function Call
cout << minNumberOfRabbits(arr, N) << endl;
return 0;
}
// This code is contributed by divyeshrabadiya07 Java
// Java program for the above approach
import java.util.*;
class GFG {
// Function to find the minimun
// number of rabbits in the forest
public static int minNumberOfRabbits(int[] answers,
int N)
{
// Initialize map
Map map
= new HashMap();
// Traverse array and map arr[i]
// to the number of occurences
for (int a : answers) {
map.put(a, map.getOrDefault(a, 0) + 1);
}
// Intialize count as 0;
int count = 0;
// Find the number groups and
// no. of rabbits in each group
for (int a : map.keySet()) {
int x = a;
int y = map.get(a);
// Find number of groups and
// multiply them with number
// of rabbits in each group
if (y % (x + 1) == 0) {
count = count + (y / (x + 1)) * (x + 1);
}
else
count
= count + ((y / (x + 1)) + 1) * (x + 1);
}
// count gives minimum number
// of rabbits in the forest
return count;
}
// Driver Code
public static void main(String[] args)
{
int arr[] = { 2, 2, 0 };
int N = arr.length;
// Function Call
System.out.println(minNumberOfRabbits(arr, N));
}
} Python3
# Python3 program for the above approach
# Function to find the minimun
# number of rabbits in the forest
def minNumberOfRabbits(answers, N):
# Initialize map
Map = {}
# Traverse array and map arr[i]
# to the number of occurences
for a in range(N):
if answers[a] in Map:
Map[answers[a]] += 1
else:
Map[answers[a]] = 1
# Intialize count as 0;
count = 0
# Find the number groups and
# no. of rabbits in each group
for a in Map:
x = a
y = Map[a]
# Find number of groups and
# multiply them with number
# of rabbits in each group
if (y % (x + 1) == 0):
count = count + (y // (x + 1)) * (x + 1)
else:
count = count + ((y // (x + 1)) + 1) * (x + 1)
# count gives minimum number
# of rabbits in the forest
return count
# Driver code
arr = [2, 2, 0]
N = len(arr)
# Function Call
print(minNumberOfRabbits(arr, N))
# This code is contributed by divyesh072019C#
// C# program for the above approach
using System;
using System.Collections.Generic;
using System.Linq;
class GFG {
// Function to find the minimun
// number of rabbits in the forest
public static int minNumberOfRabbits(int[] answers,
int N)
{
// Initialize map
Dictionary map
= new Dictionary();
// Traverse array and map arr[i]
// to the number of occurences
for (int a = 0; a < N; a++) {
if (map.ContainsKey(answers[a]))
map[answers[a]] += 1;
else
map.Add(answers[a], 1);
}
// Intialize count as 0;
int count = 0;
// Find the number groups and
// no. of rabbits in each group
for (int a = 0; a < map.Count; a++) {
int x = map.Keys.ElementAt(a);
int y = map[x];
// Find number of groups and
// multiply them with number
// of rabbits in each group
if (y % (x + 1) == 0) {
count = count + (y / (x + 1)) * (x + 1);
}
else
count
= count + ((y / (x + 1)) + 1) * (x + 1);
}
// count gives minimum number
// of rabbits in the forest
return count;
}
// Driver Code
public static void Main(string[] args)
{
int[] arr = { 2, 2, 0 };
int N = arr.Length;
// Function Call
Console.WriteLine(minNumberOfRabbits(arr, N));
}
}
// This code is contributed by AnkitRai01 Javascript
C++
// C++ program for the above approach
#include
#include
using namespace std;
// Function to find the minimun
// number of rabbits in the forest
int minNumberOfRabbits(int answers[], int n)
{
// Initialize cnt variable
int count = 0;
// Initialize map
unordered_map mp;
for (int i = 0; i < n; ++i) {
// Add to the count if not found or
// has exhausted the count of all the
// number of rabbits of same colour
if (mp[answers[i]] == 0) {
count += answers[i] + 1;
mp[answers[i]] = answers[i] + 1;
}
mp[answers[i]]--;
}
// count gives minimum number
// of rabbits in the forest
return count;
}
// Driver Code
int main()
{
int arr[] = { 10, 10, 0 };
int N = sizeof(arr) / sizeof(arr[0]);
// Function Call
cout << minNumberOfRabbits(arr, N) << endl;
return 0;
}
// This code is contributed by Bhavna Soni - bhavna23 输出
4时间复杂度： O(N)
辅助空间： O(N)
另一种解决方案是使用计数的 HasMap 并在每次遇到相同的答案 [i]时减少计数。如果相同的answers[i]再次出现并且计数为零，用answers[i]+1重置计数并添加到 answer 变量。以下是步骤：
- 用零初始化变量计数。
- 使用无序映射 mp。
- 遍历数组并执行以下操作：
- 如果answers[i]设置为零，设置mp[answers[i]] = answers[i]+1和count=count+answers[i]+1
- 最后从地图中减去计数， mp[answers[i]]– ；
- 返回 cnt 作为答案。
C++
// C++ program for the above approach
#include
#include
using namespace std;
// Function to find the minimun
// number of rabbits in the forest
int minNumberOfRabbits(int answers[], int n)
{
// Initialize cnt variable
int count = 0;
// Initialize map
unordered_map mp;
for (int i = 0; i < n; ++i) {
// Add to the count if not found or
// has exhausted the count of all the
// number of rabbits of same colour
if (mp[answers[i]] == 0) {
count += answers[i] + 1;
mp[answers[i]] = answers[i] + 1;
}
mp[answers[i]]--;
}
// count gives minimum number
// of rabbits in the forest
return count;
}
// Driver Code
int main()
{
int arr[] = { 10, 10, 0 };
int N = sizeof(arr) / sizeof(arr[0]);
// Function Call
cout << minNumberOfRabbits(arr, N) << endl;
return 0;
}
// This code is contributed by Bhavna Soni - bhavna23
输出
4时间复杂度： O(N)
辅助空间： O(N)
如果您想与行业专家一起参加直播课程，请参阅Geeks Classes Live