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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++)
}
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
}
}