Equal Beauty CodeChef SnackDown 2021 Round 1A Question The beauty of an (non-empty) array of integers is defined as the difference between its largest and smallest element. For example, the beauty of the array [2,3,4,4,6] is 6−2=4. An array A is said to be good if it is possible to partition the elements of A into two non-empty arrays B1 and B2 such that B1 and B2 have the same beauty. Each element of array A should be in exactly one array: either in B1 or in B2. For example, the array [6,2,4,4,4] is good because its elements can be partitioned into two arrays B1=[6,4,4] and B2=[2,4], where both B1 and B2 have the same beauty (6−4=4−2=2). You are given an array A of length N. In one move you can: Select an index i (1≤i≤N) and either increase Ai by 1 or decrease Ai by 1. Find the minimum number of moves required to make the array A good. Input Format The first line of input contains a single integer T, denoting the number of test cases. The description of T test cases follow. Each test
Find All Duplicates in an Array LeetCode Solution
Question
Given an integer array nums
of length n
where all the integers of nums
are in the range [1, n]
and each integer appears once or twice, return an array of all the integers that appears twice.
You must write an algorithm that runs in O(n)
time and uses only constant extra space.
Example 1:
Input: nums = [4,3,2,7,8,2,3,1] Output: [2,3]
Example 2:
Input: nums = [1,1,2] Output: [1]
Example 3:
Input: nums = [1] Output: []
Constraints:
n == nums.length
1 <= n <= 105
1 <= nums[i] <= n
- Each element in
nums
appears once or twice.
Explanation
This problem is easy and very simple to understand. No prerequisite is required to solve this problem. Since we need to solve it in constant space complexity so we had only one way out that is use the space of the given array and manipulate it. Since the elements in the array are in range [1,...,n] we can consider the elements as indexes to be precise ' element - 1 ', as it is 0-indexed array. Since we cannot use any extra space so let us mark the elements encountered in the given array only. This can be done by multiplying the elements with -1. Now for each element we will treat it as a index and check whether the value present at that index is visited or not. If it is visited then it is a duplicate else we mark it visited by multiplying with -1. That's it now code it. This is very easy to code and the code is very short. Try it yourself first then if you are stuck then visit the solution given below in Java, C++, Python. Hope you understood the explanation well and have no problem in coding it out.
Program Code
Java
class Solution {
public List<Integer> findDuplicates(int[] nums) {
List<Integer> list = new ArrayList<>();
for(int i =0;i<nums.length;i++)
{
int index = Math.abs(nums[i]) - 1;
if(nums[index]<0)list.add(index+1);
nums[index] = -nums[index];
}
return list;
}
}
C++
public:
vector<int> findDuplicates(vector<int>& nums) {
vector<int>ans;
for(int i=0;i<nums.size();i++){
nums[abs(nums[i])-1]=-nums[abs(nums[i])-1];
if(nums[abs(nums[i])-1]>0)
ans.push_back(abs(nums[i]));
}
return ans;
}
};
Python 3
class Solution:
def findDuplicates(self, nums: List[int]) -> List[int]:
return [num for num,count in Counter(nums).items() if count == 2]
More information about LeetCode Daily Challenge
More Information About LeetCode Daily Challenge
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