Counting Pairs (x, y) in an Array Where x^y > y^xIn this tutorial, we will write the Python program to find the number of pairs such that x^y>y^x. We have given two arrays X[] and Y[] containing positive integers, we need to determine the count of pairs (x, y) where x is an element from X[] and y is an element from Y[], and the condition x^y > y^x is satisfied. Example 5: Input: Output 3 Explanation: - 3^1 > 1^3 (3^1 > 1^3) - 4^5 > 5^4 (4^5 > 5^4) - 2^1 > 1^2 (2^1 > 1^2) Solution - 1:Let's understand the code snippet - Example - Output 3 Explanation - Let's understand the following explanation -
Time Complexity: O(M*N) where M and N are sizes of given arrays. Auxiliary Space: O(1) Solution - 2:Following is the more efficient approach to solve the problem Example - Output 3 Explanation - In the above code, first we imported the bisect module and defined count_pairs_with_condition() function which calculates the count of pairs satisfying the condition for a given x. If x is 0, it returns 0, as there cannot be any value in Y that satisfies the condition. If x is 1, it returns the count of zeroes in Y since any number raised to the power of 0 is 1, and therefore, x^0 is always greater than y^x for x greater than 1. It sorts the array Y and uses the bisect.bisect_right() function to find the index where x should be inserted in the sorted array. This index represents the count of elements in Y greater than x. Then we defined the count_pairs_satisfying_condition() function This function calculates the total count of pairs satisfying the condition for all values in array X. It initializes an array counts_of_Y to store counts of elements in Y with values less than 5. These counts help in counting pairs efficiently. It sorts the array Y to enable binary search in the next step. It initializes total_pairs to keep track of the total count of pairs. Time Complexity: O((n + m) * log(n + m)), where `n` is the length of array X and m is the length of array Y, and the primary factor affecting the time complexity is the sorting of both arrays, which takes O((n + m) * log(n + m)) time. Auxiliary Space: O(n + m), the additional space used by the code for arrays counts_of_Y and the sorted Y. ConclusionIn this tutorial, we explored how to find the number of pairs (x, y) in two arrays such that x^y > y^x. We provided a step-by-step explanation of the solution, including a more efficient approach that optimizes the counting process. The efficient solution uses sorting and clever counting to reduce time complexity. It's a valuable technique for solving similar problems efficiently. By understanding these methods, you can tackle such challenges effectively in your programming knowledge. |
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