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New Problem Solution - "1814. Count Nice Pairs in an Array"
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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 | ||
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/***************************************************************************************************** | ||
* | ||
* 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 | ||
******************************************************************************************************/ | ||
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class Solution { | ||
private: | ||
int rev(int n) { | ||
int x = 0; | ||
while(n > 0) { | ||
x = x*10 + (n % 10); | ||
n /= 10; | ||
} | ||
return x; | ||
} | ||
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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' | ||
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unordered_map<int, int> stat; | ||
for(auto& n : nums) { | ||
stat[n-rev(n)]++; | ||
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} | ||
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//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; | ||
} | ||
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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 ; | ||
} | ||
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return result ; | ||
} | ||
}; |