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Formatted question description: https://leetcode.ca/all/1313.html

1313. Decompress Run-Length Encoded List (Easy)

We are given a list nums of integers representing a list compressed with run-length encoding.

Consider each adjacent pair of elements [a, b] = [nums[2*i], nums[2*i+1]] (with i >= 0).  For each such pair, there are a elements with value b in the decompressed list.

Return the decompressed list.

 

Example 1:

Input: nums = [1,2,3,4]
Output: [2,4,4,4]
Explanation: The first pair [1,2] means we have freq = 1 and val = 2 so we generate the array [2].
The second pair [3,4] means we have freq = 3 and val = 4 so we generate [4,4,4].
At the end the concatenation [2] + [4,4,4,4] is [2,4,4,4].

 

Constraints:

  • 2 <= nums.length <= 100
  • nums.length % 2 == 0
  • 1 <= nums[i] <= 100

Related Topics:
Array

Similar Questions:

Solution 1.

  • class Solution {
        public int[] decompressRLElist(int[] nums) {
            List<Integer> list = new ArrayList<Integer>();
            int length = nums.length;
            for (int i = 0; i < length; i += 2) {
                int count = nums[i];
                int value = nums[i + 1];
                for (int j = 0; j < count; j++)
                    list.add(value);
            }
            int size = list.size();
            int[] decompressed = new int[size];
            for (int i = 0; i < size; i++)
                decompressed[i] = list.get(i);
            return decompressed;
        }
    }
    
    ############
    
    class Solution {
        public int[] decompressRLElist(int[] nums) {
            int n = 0;
            for (int i = 0; i < nums.length; i += 2) {
                n += nums[i];
            }
            int[] res = new int[n];
            for (int i = 1, k = 0; i < nums.length; i += 2) {
                for (int j = 0; j < nums[i - 1]; ++j) {
                    res[k++] = nums[i];
                }
            }
            return res;
        }
    }
    
  • // OJ: https://leetcode.com/problems/decompress-run-length-encoded-list/
    // Time: O(N) where N is the length of the output array.
    // Space: O(1)
    class Solution {
    public:
        vector<int> decompressRLElist(vector<int>& nums) {
            vector<int> ans;
            for (int i = 0; i < nums.size(); i += 2) {
                for (int j = 0; j < nums[i]; ++j) ans.push_back(nums[i + 1]);
            }
            return ans;
        }
    };
    
  • class Solution:
        def decompressRLElist(self, nums: List[int]) -> List[int]:
            res = []
            for i in range(1, len(nums), 2):
                res.extend([nums[i]] * nums[i - 1])
            return res
    
    
    
  • func decompressRLElist(nums []int) []int {
    	var res []int
    	for i := 1; i < len(nums); i += 2 {
    		for j := 0; j < nums[i-1]; j++ {
    			res = append(res, nums[i])
    		}
    	}
    	return res
    }
    
  • function decompressRLElist(nums: number[]): number[] {
        let n = nums.length >> 1;
        let ans = [];
        for (let i = 0; i < n; i++) {
            let freq = nums[2 * i],
                val = nums[2 * i + 1];
            ans.push(...new Array(freq).fill(val));
        }
        return ans;
    }
    
    
  • impl Solution {
        pub fn decompress_rl_elist(nums: Vec<i32>) -> Vec<i32> {
            let n = nums.len() >> 1;
            let mut ans = Vec::new();
            for i in 0..n {
                for _ in 0..nums[2 * i] {
                    ans.push(nums[2 * i + 1]);
                }
            }
            ans
        }
    }
    
    

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