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1313. Decompress Run-Length Encoded List

Description

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

Consider each adjacent pair of elements [freq, val] = [nums[2*i], nums[2*i+1]] (with i >= 0).  For each such pair, there are freq elements with value val concatenated in a sublist. Concatenate all the sublists from left to right to generate 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] is [2,4,4,4].

Example 2:

Input: nums = [1,1,2,3]
Output: [1,3,3]

 

Constraints:

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

Solutions

Solution 1: Simulation

We can directly simulate the process described in the problem. Traverse the array $\textit{nums}$ from left to right, each time taking out two numbers $\textit{freq}$ and $\textit{val}$, then repeat $\textit{val}$ $\textit{freq}$ times, and add these $\textit{freq}$ $\textit{val}$s to the answer array.

The time complexity is $O(n)$, where $n$ is the length of the array $\textit{nums}$. We only need to traverse the array $\textit{nums}$ once. Ignoring the space consumption of the answer array, the space complexity is $O(1)$.

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class Solution:
    def decompressRLElist(self, nums: List[int]) -> List[int]:
        return [nums[i + 1] for i in range(0, len(nums), 2) for _ in range(nums[i])]

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class Solution {
    public int[] decompressRLElist(int[] nums) {
        List<Integer> ans = new ArrayList<>();
        for (int i = 0; i < nums.length; i += 2) {
            for (int j = 0; j < nums[i]; ++j) {
                ans.add(nums[i + 1]);
            }
        }
        return ans.stream().mapToInt(i -> i).toArray();
    }
}

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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;
    }
};

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func decompressRLElist(nums []int) (ans []int) {
    for i := 1; i < len(nums); i += 2 {
        for j := 0; j < nums[i-1]; j++ {
            ans = append(ans, nums[i])
        }
    }
    return
}

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function decompressRLElist(nums: number[]): number[] {
    const ans: number[] = [];
    for (let i = 0; i < nums.length; i += 2) {
        for (let j = 0; j < nums[i]; j++) {
            ans.push(nums[i + 1]);
        }
    }
    return ans;
}

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impl Solution {
    pub fn decompress_rl_elist(nums: Vec<i32>) -> Vec<i32> {
        let mut ans = Vec::new();
        let n = nums.len();
        let mut i = 0;
        while i < n {
            let freq = nums[i];
            let val = nums[i + 1];
            for _ in 0..freq {
                ans.push(val);
            }
            i += 2;
        }
        ans
    }
}

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/**
 * Note: The returned array must be malloced, assume caller calls free().
 */
int* decompressRLElist(int* nums, int numsSize, int* returnSize) {
    int n = 0;
    for (int i = 0; i < numsSize; i += 2) {
        n += nums[i];
    }
    int* ans = (int*) malloc(n * sizeof(int));
    *returnSize = n;
    int k = 0;
    for (int i = 0; i < numsSize; i += 2) {
        int freq = nums[i];
        int val = nums[i + 1];
        for (int j = 0; j < freq; j++) {
            ans[k++] = val;
        }
    }
    return ans;
}

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