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961_N-Repeated_Element_in_Size_2N_Array
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README.md

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| 946 | [Validate Stack Sequences](https://leetcode.com/problems/validate-stack-sequences/) | [Python](https://github.com/qiyuangong/leetcode/blob/master/python/946_Validate_Stack_Sequences.py) [Java](https://github.com/qiyuangong/leetcode/blob/master/java/946_Validate_Stack_Sequences.java) | Add a stack named inStack to help going through pushed and popped. O(n) and O(n) |
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| 953 | [Verifying an Alien Dictionary](https://leetcode.com/contest/weekly-contest-114/problems/verifying-an-alien-dictionary/) | [Python](https://github.com/qiyuangong/leetcode/blob/master/python/953_Verifying_an_Alien_Dictionary.py) [Java](https://github.com/qiyuangong/leetcode/blob/master/java/953_Verifying_an_Alien_Dictionary.java) | Use hashmap to store index of each value, then create a comparator based on this index, O(n) and O(n) |
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| 954 | [Array of Doubled Pairs](https://leetcode.com/contest/weekly-contest-114/problems/array-of-doubled-pairs/) | [Python](https://github.com/qiyuangong/leetcode/blob/master/python/954_Array_of_Doubled_Pairs.py) [Java](https://github.com/qiyuangong/leetcode/blob/master/java/954_Array_of_Doubled_Pairs.java) | Sort, then use hashmap to store the frequency of each value. Then, check n, 2 * n in hashmap, O(nlogn) and O(n) |
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| 961 | [N-Repeated Element in Size 2N Array](https://leetcode.com/problems/n-repeated-element-in-size-2n-array/submissions/) | [Python](https://github.com/qiyuangong/leetcode/blob/master/python/961_N-Repeated_Element_in_Size_2N_Array.py) [Java](https://github.com/qiyuangong/leetcode/blob/master/java/961_N-Repeated_Element_in_Size_2N_Array.java) | Hash and count number, O(n) and O(n) |
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| # | To Understand |
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class Solution {
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public int repeatedNTimes(int[] A) {
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HashMap<Integer, Integer> hash = new HashMap<>();
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int ans = A[0];
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for (int n: A) {
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int count = hash.getOrDefault(n, 0) + 1;
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hash.put(n, count);
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if (count >= hash.get(ans)) ans = n;
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}
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return ans;
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}
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}
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import collections
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class Solution(object):
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def repeatedNTimes(self, A):
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"""
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:type A: List[int]
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:rtype: int
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"""
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counter = collections.Counter(A)
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return counter.most_common(1)[0][0]
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if __name__ == '__main__':
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s = Solution()
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print s.repeatedNTimes([1, 2, 3, 3])
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print s.repeatedNTimes([2, 1, 2, 5, 3, 2])
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print s.repeatedNTimes([5, 1, 5, 2, 5, 3, 5, 4])

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