In this guide, you will get 2354. Number of Excellent Pairs LeetCode Solution with the best time and space complexity. The solution to Number of Excellent Pairs problem is provided in various programming languages like C++, Java, and Python. This will be helpful for you if you are preparing for placements, hackathons, interviews, or practice purposes. The solutions provided here are very easy to follow and include detailed explanations.
You are given a 0-indexed positive integer array nums and a positive integer k.
A pair of numbers (num1, num2) is called excellent if the following conditions are satisfied:
Both the numbers num1 and num2 exist in the array nums.
The sum of the number of set bits in num1 OR num2 and num1 AND num2 is greater than or equal to k, where OR is the bitwise OR operation and AND is the bitwise AND operation.
Return the number of distinct excellent pairs.
Two pairs (a, b) and (c, d) are considered distinct if either a != c or b != d. For example, (1, 2) and (2, 1) are distinct.
Note that a pair (num1, num2) such that num1 == num2 can also be excellent if you have at least one occurrence of num1 in the array.
Example 1:
Input: nums = [1,2,3,1], k = 3
Output: 5
Explanation: The excellent pairs are the following:
– (3, 3). (3 AND 3) and (3 OR 3) are both equal to (11) in binary. The total number of set bits is 2 + 2 = 4, which is greater than or equal to k = 3.
– (2, 3) and (3, 2). (2 AND 3) is equal to (10) in binary, and (2 OR 3) is equal to (11) in binary. The total number of set bits is 1 + 2 = 3.
– (1, 3) and (3, 1). (1 AND 3) is equal to (01) in binary, and (1 OR 3) is equal to (11) in binary. The total number of set bits is 1 + 2 = 3.
So the number of excellent pairs is 5.
Example 2:
2354. Number of Excellent Pairs LeetCode Solution in C++
class Solution {
public:
long long countExcellentPairs(vector<int>& nums, int k) {
constexpr int kMaxBit = 30;
// bits(num1 | num2) + bits(num1 & num2) = bits(num1) + bits(num2)
long ans = 0;
vector<long> count(kMaxBit);
for (const unsigned num : unordered_set<int>(nums.begin(), nums.end()))
++count[popcount(num)];
for (int i = 0; i < kMaxBit; ++i)
for (int j = 0; j < kMaxBit; ++j)
if (i + j >= k)
ans += count[i] * count[j];
return ans;
}
}; /* code provided by PROGIEZ */
2354. Number of Excellent Pairs LeetCode Solution in Java
class Solution {
public long countExcellentPairs(int[] nums, int k) {
final int kMaxBit = 30;
// bits(num1 | num2) + bits(num1 & num2) = bits(num1) + bits(num2)
long ans = 0;
long[] count = new long[kMaxBit];
for (final int num : Arrays.stream(nums).boxed().collect(Collectors.toSet()))
++count[Integer.bitCount(num)];
for (int i = 0; i < kMaxBit; ++i)
for (int j = 0; j < kMaxBit; ++j)
if (i + j >= k)
ans += count[i] * count[j];
return ans;
}
} // code provided by PROGIEZ
2354. Number of Excellent Pairs LeetCode Solution in Python
class Solution:
def countExcellentPairs(self, nums: list[int], k: int) -> int:
count = collections.Counter(map(int.bit_count, set(nums)))
return sum(count[i] * count[j]
for i in count
for j in count
if i + j >= k) # code by PROGIEZ
