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MaximumErasureValue.cpp
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MaximumErasureValue.cpp
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// Source : https://leetcode.com/problems/maximum-erasure-value/
// Author : Hao Chen
// Date : 2021-05-07
/*****************************************************************************************************
*
* You are given an array of positive integers nums and want to erase a subarray containing unique
* elements. The score you get by erasing the subarray is equal to the sum of its elements.
*
* Return the maximum score you can get by erasing exactly one subarray.
*
* An array b is called to be a subarray of a if it forms a contiguous subsequence of a, that is, if
* it is equal to a[l],a[l+1],...,a[r] for some (l,r).
*
* Example 1:
*
* Input: nums = [4,2,4,5,6]
* Output: 17
* Explanation: The optimal subarray here is [2,4,5,6].
*
* Example 2:
*
* Input: nums = [5,2,1,2,5,2,1,2,5]
* Output: 8
* Explanation: The optimal subarray here is [5,2,1] or [1,2,5].
*
* Constraints:
*
* 1 <= nums.length <= 10^5
* 1 <= nums[i] <= 10^4
******************************************************************************************************/
class Solution {
public:
int maximumUniqueSubarray(vector<int>& nums) {
//unordered_map<int, int> pos;
const int NIL = -1;
int pos[10001];
memset(pos, NIL, sizeof(pos));
int start=0;
int max_sum =0, sum = 0;
for(int i = 0; i < nums.size(); i++) {
int n = nums[i];
// if find duplicated number
if ( pos[n] != NIL) {
max_sum = max(max_sum, sum);
//remove the previous numbers until to duplicatied position
for(;start <= pos[n]; start++){
sum -= nums[start];
pos[nums[start]] = NIL;
}
}
sum += n;
pos[n] = i;
}
max_sum = max( max_sum , sum );
return max_sum;
}
};