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 109 + 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 <= 105
0 <= nums[i] <= 109
Solution: Two Sum
Key = x – rev(x)
Time complexity: O(n)
Space complexity: O(n)
C++
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// Author: Huahua class Solution { public: int countNicePairs(vector<int>& nums) { constexpr int kMod = 1e9 + 7; unordered_map<int, int> m; long ans = 0; for (int x : nums) { string s = to_string(x); reverse(begin(s), end(s)); ans += m[x - stoi(s)]++; } return ans % kMod; } }; |
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