You are given a 0-indexed binary array nums
of length n
. nums
can be divided at index i
(where 0 <= i <= n)
into two arrays (possibly empty) numsleft
and numsright
:
numsleft
has all the elements ofnums
between index0
andi - 1
(inclusive), whilenumsright
has all the elements of nums between indexi
andn - 1
(inclusive).- If
i == 0
,numsleft
is empty, whilenumsright
has all the elements ofnums
. - If
i == n
,numsleft
has all the elements of nums, whilenumsright
is empty.
The division score of an index i
is the sum of the number of 0
‘s in numsleft
and the number of 1
‘s in numsright
.
Return all distinct indices that have the highest possible division score. You may return the answer in any order.
Example 1:
Input: nums = [0,0,1,0] Output: [2,4] Explanation: Division at index - 0: numsleft is []. numsright is [0,0,1,0]. The score is 0 + 1 = 1. - 1: numsleft is [0]. numsright is [0,1,0]. The score is 1 + 1 = 2. - 2: numsleft is [0,0]. numsright is [1,0]. The score is 2 + 1 = 3. - 3: numsleft is [0,0,1]. numsright is [0]. The score is 2 + 0 = 2. - 4: numsleft is [0,0,1,0]. numsright is []. The score is 3 + 0 = 3. Indices 2 and 4 both have the highest possible division score 3. Note the answer [4,2] would also be accepted.
Example 2:
Input: nums = [0,0,0] Output: [3] Explanation: Division at index - 0: numsleft is []. numsright is [0,0,0]. The score is 0 + 0 = 0. - 1: numsleft is [0]. numsright is [0,0]. The score is 1 + 0 = 1. - 2: numsleft is [0,0]. numsright is [0]. The score is 2 + 0 = 2. - 3: numsleft is [0,0,0]. numsright is []. The score is 3 + 0 = 3. Only index 3 has the highest possible division score 3.
Example 3:
Input: nums = [1,1] Output: [0] Explanation: Division at index - 0: numsleft is []. numsright is [1,1]. The score is 0 + 2 = 2. - 1: numsleft is [1]. numsright is [1]. The score is 0 + 1 = 1. - 2: numsleft is [1,1]. numsright is []. The score is 0 + 0 = 0. Only index 0 has the highest possible division score 2.
Constraints:
n == nums.length
1 <= n <= 105
nums[i]
is either0
or1
.
Solution: Precompute + Prefix Sum
Count how many ones in the array, track the prefix sum to compute score for each index in O(1).
Time complexity: O(n)
Space complexity: O(n)
C++
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// Author: Huahua class Solution { public: vector<int> maxScoreIndices(vector<int>& nums) { const int n = nums.size(); const int t = accumulate(begin(nums), end(nums), 0); vector<int> ans; for (int i = 0, p = 0, b = 0; i <= n; ++i) { const int s = (i - p) + (t - p); if (s > b) { b = s; ans.clear(); } if (s == b) ans.push_back(i); if (i != n) p += nums[i]; } return ans; } }; |
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