Build Array from Permutation Leetcode Solution

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Build Array from Permutation

Given a zero-based permutation nums (0-indexed), build an array ans of the same length where ans[i] = nums[nums[i]] for each 0 <= i < nums.length and return it.

See Also: Weekly Contest 248

zero-based permutation nums is an array of distinct integers from 0 to nums.length - 1 (inclusive).

Example 1:

Input: nums = [0,2,1,5,3,4]
Output: [0,1,2,4,5,3]
Explanation: The array ans is built as follows: 
ans = [nums[nums[0]], nums[nums[1]], nums[nums[2]], nums[nums[3]], nums[nums[4]], nums[nums[5]]]
    = [nums[0], nums[2], nums[1], nums[5], nums[3], nums[4]]
    = [0,1,2,4,5,3]

Example 2:

Input: nums = [5,0,1,2,3,4]
Output: [4,5,0,1,2,3]
Explanation: The array ans is built as follows:
ans = [nums[nums[0]], nums[nums[1]], nums[nums[2]], nums[nums[3]], nums[nums[4]], nums[nums[5]]]
    = [nums[5], nums[0], nums[1], nums[2], nums[3], nums[4]]
    = [4,5,0,1,2,3]

Constraints:

  • 1 <= nums.length <= 1000
  • 0 <= nums[i] < nums.length
  • The elements in nums are distinct.

Follow-up: Can you solve it without using an extra space (i.e., O(1) memory)?

Solution

Program C++

class Solution
{
public:
vector buildArray(vector& nums)
{
int n=nums.size();
vector v(n,0);
for(int i=0;i<n;i++)
{
v[i]=nums[nums[i]];
}
return v;
}
};

Program Python

class Solution:
    def buildArray(self, nums: List[int]) -> List[int]:
        ans = [None]*len(nums)
        for i in range(len(nums)):
            ans[i] = nums[nums[i]]
        return ans

Program Java

  class Solution {
    public int[] buildArray(int[] nums) {
         int size=nums.length;
        int[] a=new int[size];
         
        for (int i = 0; i < size; i++) {
	        int product = (nums[nums[i]] % size) * size + nums[i];
	        a[i]=product;
	    }

	    for (int i = 0; i < size; i++) {
	        int temp = a[i];
	        a[i]=temp/size;
	    } 
        return a;
    }
} ```

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