In the Rabbits in Forest problem, we are given an array that contains some rabbits’ answers to the question, \u201cHow many rabbits have the same color as yours\u201d. We have to count the minimum number of rabbits present in the forest.<\/strong> <\/p>\n\n\n\n Note that the Array size may be less than the total number of rabbits. Let\u2019s understand this with an example:<\/p>\n\n\n\n Input:<\/strong> [0, 2, 1, 1]<\/p>\n\n\n\n Output:<\/strong> 6<\/p>\n\n\n\n The answer will look like this:<\/p>\n\n\n\n Another similar question about removing duplicate elements from an array, a 4-sum problem<\/a>, etc., using a map. <\/p>\n\n\n\n Let’s assume you have an array [1, 1, 1, 1], which means you have 2 rabbits of the same color white, and 2 rabbits of the same color green. This can also be observed by finding the number of rabbits with the same value as a[i]. <\/p>\n\n\n\n Now, find their remainder when you divide this count by a[i]+1. Add this remainder to the count. Then, add the count to the answer variable.<\/p>\n\n\n\n Let’s take an example:<\/p>\n\n\n\n The time complexity for the above process is O( n log(n) ) as we are storing the value into the map which requires log(n) time complexity. We are iterating over the loop that took O(n) time complexity.\u00a0The space complexity will be O(n) as we are storing values on the map.<\/p>\n\n\n\n You can also find the intersection of two arrays<\/a> using hashmap.<\/p>\n\n\n\n As mentioned above, we are using O(n) space complexity to store all the values in the array. This can be reduced if we sort the array and count the values till they are the same and then find the remainder. In that case, we do not have to store the values in the map and our space will reduce to O(1).<\/p>\n\n\n\n Let\u2019s examine this approach:<\/p>\n\n\n\n Let\u2019s examine the code for the above process.<\/p>\n\n\n\n Here is the solution implemented in C++ for Rabbits in Forest problem:<\/p>\n\n\n\n We can program it with Java also:<\/p>\n\n\n\n Below is the Python program for the sorting approach:<\/p>\n\n\n\n Output:<\/strong><\/p>\n\n\n\n The time complexity for the Sorting approach to solving the Rabbits in Forest problem is O( n log(n) ).<\/strong> We are iterating over a loop whose time complexity does O ( n ) so overall complexity becomes O( n log(n) ).\u00a0Space complexity will be reduced to O(1) as no extra space is used.\u00a0<\/p>\n\n\n\n We can compare both the approaches to see how we optimized it:<\/p>\n\n\n\n![\"Initial](\"https:\/\/lh7-us.googleusercontent.com\/CvJkUwLjufvzAFHZqE5ktI8rcbVsB73XMss_yHRTzRcHx_qVyAh0cT3C9oHa5t1zwfW7Ehv6YRlCrUCPWtr6IPyV8YA2IRaILxwKrOb145IiHL9UQVfhTG_gTfBPujwifxcWCAHEf-lvKeg1hKexag0\")
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![\"Result](\"https:\/\/lh7-us.googleusercontent.com\/C0EiXcJFT7spTyYjw2mkUdxMjI7e5BtL8-X84vI06AXi3LZHDFr-I0VeamqeygMAFIkmICCMiQJ-Lf-iXy0j4dNgfQhRjA81KgYjzHe8MOHdbYuoj9TGpsZhQWPeRx7gTmHS6vO2bfcwhWN9T9FXaXA\")
Hash Map Approach<\/strong><\/h2>\n\n\n\n
![\"Hash](\"https:\/\/lh7-us.googleusercontent.com\/Q74gFvp2ypDQvI3DYAl-WXXzZI9cP_kHuyYR1BlBJnSjxGhFvnGD725A5KIoROyCRluiewg132T_Q0aTtriKSwlZpO7kEMvFPqIZ53pOgc6fjSbX4cN5jiewyVfGqeWaQm95xQw0wVbmYMNMuxhgz6s\")
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Optimize Sorting<\/strong> Approach<\/strong><\/h2>\n\n\n\n
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C++ Code<\/strong><\/h3>\n\n\n\n
#include<bits\/stdc++.h>\nusing namespace std;\n\nint main(){\n\n vector<int> arr = {1, 1, 2, 0, 1};\n \/\/ sort the given array\n sort(arr.begin(), arr.end());\n\n \/\/ count variable for count the frequency of the same elements.\n int count = 1;\n \/\/ ans variable which stores total rabbits and n stores the size of the array.\n int ans = 0, n = arr.size();\n\n for(int i=0; i<arr.size(); i++ ){\n \n if(i == arr.size()-1 || (arr[i]!=arr[i+1])){\n \/\/ find remainder.\n int rem = count%(arr[i]+1);\n if(rem) ans = ans + (arr[i]+1) - rem;\n count = 1;\n }else{\n \/\/ if current and next values are not same increment our count variable.\n count = count + 1;\n }\n }\n \n ans = ans + n;\n cout<<ans<<endl;\n return 0;\n}<\/pre><\/div>\n\n\n\nJava Code<\/strong><\/h3>\n\n\n\n
import java.util.*;\n\nclass RabbitsInForest {\n public static void main(String[] args) {\n List<Integer> arr = new ArrayList<>(Arrays.asList(1, 1, 2, 0, 1));\n \n \/\/ Sort the given array\n Collections.sort(arr);\n\n \/\/ Count variable for counting the frequency of the same elements.\n int count = 1;\n \/\/ Ans variable which stores total rabbits and n stores the size of the array.\n int ans = 0;\n int n = arr.size();\n\n for (int i = 0; i < arr.size(); i++) {\n if (i == arr.size() - 1 || arr.get(i) != arr.get(i + 1)) {\n \/\/ Find remainder.\n int rem = count % (arr.get(i) + 1);\n if (rem != 0) ans += (arr.get(i) + 1) - rem;\n count = 1;\n } else {\n \/\/ If current and next values are not same increment our count variable.\n count++;\n }\n }\n\n ans += n;\n System.out.println(ans);\n }\n}<\/pre><\/div>\n\n\n\nPython Code<\/strong><\/h3>\n\n\n\n
arr = [1, 1, 2, 0, 1]\n\n# Sort the given array\narr.sort()\n\n# Count variable for counting the frequency of the same elements.\ncount = 1\n# Ans variable which stores total rabbits and n stores array size.\nans = 0\nn = len(arr)\n\nfor i in range(len(arr)):\n if i == len(arr) - 1 or arr[i] != arr[i + 1]:\n # Find remainder.\n rem = count % (arr[i] + 1)\n if rem:\n ans += (arr[i] + 1) - rem\n count = 1\n else:\n # If current and next values are not same increment our count variable.\n count += 1\n\nans += n\nprint(ans)<\/pre><\/div>\n\n\n\n
8<\/code><\/pre>\n\n\n\n