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New Problem Solution - "1814. Count Nice Pairs in an Array"
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haoel committed Apr 6, 2021
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Expand Up @@ -13,6 +13,7 @@ LeetCode
|1818|[Minimum Absolute Sum Difference](https://leetcode.com/problems/minimum-absolute-sum-difference/) | [C++](./algorithms/cpp/minimumAbsoluteSumDifference/MinimumAbsoluteSumDifference.cpp)|Medium|
|1817|[Finding the Users Active Minutes](https://leetcode.com/problems/finding-the-users-active-minutes/) | [C++](./algorithms/cpp/findingTheUsersActiveMinutes/FindingTheUsersActiveMinutes.cpp)|Medium|
|1816|[Truncate Sentence](https://leetcode.com/problems/truncate-sentence/) | [C++](./algorithms/cpp/truncateSentence/TruncateSentence.cpp)|Easy|
|1814|[Count Nice Pairs in an Array](https://leetcode.com/problems/count-nice-pairs-in-an-array/) | [C++](./algorithms/cpp/countNicePairsInAnArray/CountNicePairsInAnArray.cpp)|Medium|
|1813|[Sentence Similarity III](https://leetcode.com/problems/sentence-similarity-iii/) | [C++](./algorithms/cpp/sentenceSimilarity/SentenceSimilarity.III.cpp)|Medium|
|1812|[Determine Color of a Chessboard Square](https://leetcode.com/problems/determine-color-of-a-chessboard-square/) | [C++](./algorithms/cpp/determineColorOfAChessboardSquare/DetermineColorOfAChessboardSquare.cpp)|Easy|
|1808|[Maximize Number of Nice Divisors](https://leetcode.com/problems/maximize-number-of-nice-divisors/) | [C++](./algorithms/cpp/maximizeNumberOfNiceDivisors/MaximizeNumberOfNiceDivisors.cpp)|Hard|
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86 changes: 86 additions & 0 deletions algorithms/cpp/countNicePairsInAnArray/CountNicePairsInAnArray.cpp
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// Source : https://leetcode.com/problems/count-nice-pairs-in-an-array/
// Author : Hao Chen
// Date : 2021-04-06

/*****************************************************************************************************
*
* You are given an array nums that consists of non-negative integers. Let us define rev(x) as the
* reverse of the non-negative integer x. For example, rev(123) = 321, and rev(120) = 21. A pair of
* indices (i, j) is nice if it satisfies all of the following conditions:
*
* 0 <= i < j < nums.length
* nums[i] + rev(nums[j]) == nums[j] + rev(nums[i])
*
* Return the number of nice pairs of indices. Since that number can be too large, return it modulo
* 10^9 + 7.
*
* Example 1:
*
* Input: nums = [42,11,1,97]
* Output: 2
* Explanation: The two pairs are:
* - (0,3) : 42 + rev(97) = 42 + 79 = 121, 97 + rev(42) = 97 + 24 = 121.
* - (1,2) : 11 + rev(1) = 11 + 1 = 12, 1 + rev(11) = 1 + 11 = 12.
*
* Example 2:
*
* Input: nums = [13,10,35,24,76]
* Output: 4
*
* Constraints:
*
* 1 <= nums.length <= 10^5
* 0 <= nums[i] <= 10^9
******************************************************************************************************/

class Solution {
private:
int rev(int n) {
int x = 0;
while(n > 0) {
x = x*10 + (n % 10);
n /= 10;
}
return x;
}

public:
int countNicePairs(vector<int>& nums) {
return countNicePairs02(nums);
return countNicePairs01(nums);
}
int countNicePairs01(vector<int>& nums) {
// suppose n' = rev(n)
// define: a + b' == b + a'
// then: a - a' == b - b'

unordered_map<int, int> stat;
for(auto& n : nums) {
stat[n-rev(n)]++;

}

//if there are n elements has same value,
// then there are n*(n-1)/2 unique pairs.
int result = 0;
for(auto& [n, cnt] : stat) {
result = (result + cnt * (cnt -1l) / 2) % 1000000007;
}
return result;
}

int countNicePairs02(vector<int>& nums) {
// suppose n' = rev(n)
// define: a + b' == b + a'
// then: a - a' == b - b'
int result = 0;
unordered_map<int, int> stat;
for(auto& n : nums) {
int delta = n-rev(n);
stat[delta]++;
result = (result + (stat[delta] - 1l)) % 1000000007 ;
}

return result ;
}
};

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