Problem
We define the conversion array conver
of an array arr
as follows:
conver[i] = arr[i] + max(arr[0..i])
wheremax(arr[0..i])
is the maximum value ofarr[j]
over0 <= j <= i
.
We also define the score of an array arr
as the sum of the values of the conversion array of arr
.
Given a 0-indexed integer array nums
of length n
, return **an array *ans
* of length n
where ans[i]
is the score of the prefix** nums[0..i]
.
Example 1:
Input: nums = [2,3,7,5,10]
Output: [4,10,24,36,56]
Explanation:
For the prefix [2], the conversion array is [4] hence the score is 4
For the prefix [2, 3], the conversion array is [4, 6] hence the score is 10
For the prefix [2, 3, 7], the conversion array is [4, 6, 14] hence the score is 24
For the prefix [2, 3, 7, 5], the conversion array is [4, 6, 14, 12] hence the score is 36
For the prefix [2, 3, 7, 5, 10], the conversion array is [4, 6, 14, 12, 20] hence the score is 56
Example 2:
Input: nums = [1,1,2,4,8,16]
Output: [2,4,8,16,32,64]
Explanation:
For the prefix [1], the conversion array is [2] hence the score is 2
For the prefix [1, 1], the conversion array is [2, 2] hence the score is 4
For the prefix [1, 1, 2], the conversion array is [2, 2, 4] hence the score is 8
For the prefix [1, 1, 2, 4], the conversion array is [2, 2, 4, 8] hence the score is 16
For the prefix [1, 1, 2, 4, 8], the conversion array is [2, 2, 4, 8, 16] hence the score is 32
For the prefix [1, 1, 2, 4, 8, 16], the conversion array is [2, 2, 4, 8, 16, 32] hence the score is 64
Constraints:
1 <= nums.length <= 105
1 <= nums[i] <= 109
Solution (Java)
class Solution {
public long[] findPrefixScore(int[] nums) {
long[] arr= new long[nums.length];
long max=nums[0];
long sum=0;
for(int i=0;i<nums.length;i++){
max=max<=nums[i]?nums[i]:max;
sum=sum+nums[i]+max;
arr[i]=sum;
}
return arr;
}
}
Explain:
nope.
Complexity:
- Time complexity : O(n).
- Space complexity : O(n).