Leetcode 2934: Minimum Operations to Maximize Last Elements in Arrays
You are given two integer arrays, nums1 and nums2, both having length n. You are allowed to perform a series of operations (possibly none). In each operation, you can select an index i in the range [0, n-1] and swap the values of nums1[i] and nums2[i]. The goal is to satisfy two conditions: the last element of nums1 is equal to the maximum value in nums1, and the last element of nums2 is equal to the maximum value in nums2. Return the minimum number of operations required, or -1 if it’s impossible to satisfy both conditions.
Problem
Approach
Steps
Complexity
Input: The input consists of two integer arrays nums1 and nums2, both having length n.
Example: nums1 = [4, 2, 5], nums2 = [1, 3, 6]
Constraints:
• 1 <= n == nums1.length == nums2.length <= 1000
• 1 <= nums1[i] <= 10^9
• 1 <= nums2[i] <= 10^9
Output: Return the minimum number of operations needed to meet both conditions or -1 if it is impossible.
Example: 1
Constraints:
• If it's impossible, return -1.
Goal: To determine the minimum number of swaps required to satisfy both conditions, or return -1 if it's not possible.
Steps:
• Iterate over the elements of nums1 and nums2 while checking the possible swaps.
• Check the max elements of both arrays and adjust their last positions.
• If no valid swap is found, return -1, else return the minimum number of swaps needed.
Goal: The solution must handle up to 1000 elements efficiently.
Steps:
• 1 <= n <= 1000
• The values in nums1 and nums2 are within the range of 1 to 10^9.
Assumptions:
• The arrays nums1 and nums2 are of equal length.
• The arrays contain positive integers.
• Input: nums1 = [4, 2, 5], nums2 = [1, 3, 6]
• Explanation: In this example, a single swap can make both arrays satisfy the condition.
• Input: nums1 = [7, 3, 5, 1], nums2 = [9, 2, 4, 6]
• Explanation: Two swaps are required to meet the conditions for both arrays.
• Input: nums1 = [4, 2, 5], nums2 = [1, 3, 6]
• Explanation: It's impossible to meet the conditions because one of the arrays is already invalid.
Approach: To solve the problem, we need to iterate through both arrays and keep track of the swaps that would result in both conditions being satisfied.
Observations:
• The maximum elements of both arrays must be placed in the last index of their respective arrays.
• Each swap must either help move the maximum element of nums1 or nums2 to the right place.
• This problem requires comparing the max elements of both arrays and identifying when a swap would be beneficial.
Steps:
• Start by checking the maximum element in each array and their current positions.
• Iterate through both arrays, counting swaps needed to move the maximum elements to their correct positions.
• Return the result based on the number of swaps or -1 if no valid solution is possible.
Empty Inputs:
• An empty input is not possible due to the constraint n >= 1.
Large Inputs:
• When handling large arrays, ensure that the algorithm operates within time limits.
Special Values:
• If arrays have all identical values, no swaps are needed.
Constraints:
• Ensure that the solution works for both arrays containing large integers up to 10^9.
int minOperations(vector<int>& A, vector<int>& B) {
int dp1 = 0, dp2 = 0, n = A.size(), mi = min(A[n - 1], B[n - 1]), ma = max(A[n - 1], B[n - 1]);
for (int i = 0; i < n; i++) {
int a = A[i], b = B[i];
if (max(a, b) > ma) return -1;
if (min(a, b) > mi) return -1;
if (a > A[n - 1] || b > B[n - 1]) dp1++;
if (a > B[n - 1] || b > A[n - 1]) dp2++;
}
return min(dp1, dp2);
}
1 : Function Definition
int minOperations(vector<int>& A, vector<int>& B) {
Defines the main function 'minOperations', which takes two integer vectors, A and B, and returns an integer representing the minimum number of operations required.
2 : Variable Initialization
int dp1 = 0, dp2 = 0, n = A.size(), mi = min(A[n - 1], B[n - 1]), ma = max(A[n - 1], B[n - 1]);
Initializes the variables: dp1 and dp2 to count operations, n for the size of arrays, mi for the minimum value between the last elements of A and B, and ma for the maximum value.
3 : Loop Setup
for (int i = 0; i < n; i++) {
Sets up a loop to iterate through each element of the arrays A and B.
4 : Element Assignment
int a = A[i], b = B[i];
Assigns the current elements of arrays A and B to variables a and b respectively.
5 : Check for Maximum Constraint
if (max(a, b) > ma) return -1;
Checks if the maximum of a and b exceeds the maximum allowed value ma, returning -1 if the condition is met.
6 : Check for Minimum Constraint
if (min(a, b) > mi) return -1;
Checks if the minimum of a and b exceeds the minimum allowed value mi, returning -1 if the condition is met.
7 : Increment dp1
if (a > A[n - 1] || b > B[n - 1]) dp1++;
Increments dp1 if either a is greater than the last element of A or b is greater than the last element of B.
8 : Increment dp2
if (a > B[n - 1] || b > A[n - 1]) dp2++;
Increments dp2 if either a is greater than the last element of B or b is greater than the last element of A.
9 : Return Statement
return min(dp1, dp2);
Returns the minimum of dp1 and dp2, representing the minimal operations needed.
Best Case: O(n)
Average Case: O(n)
Worst Case: O(n)
Description: The time complexity is linear in relation to the size of the input arrays.
Best Case: O(1)
Worst Case: O(n)
Description: The space complexity is O(n) due to the need to store temporary results.
| LeetCode Solutions Library / DSA Sheets / Course Catalog |
|---|
comments powered by Disqus