CA 06 - Find the Maximum and Minimum Element in the Array



Min and Max in Array
Input: arr[] = [1, 4, 3, 5, 8, 6]
Output: [1, 8]
Explanation: minimum and maximum elements of array are 1 and 8.Input: arr[] = [12, 3, 15, 7, 9]

Output: [3, 15]
Explanation: minimum and maximum element of array are 3 and 15.
MY CODE:
The min and max are the functions which shows the maximum and minimum element
class Solution:
   
class Solution:
    def getMinMax(self, arr):
        a = arr[0]   # min
        b = arr[0]   # max
        for c in range(1, len(arr)):
            d = arr[c]
            if d < a:
                a = d
            if d > b:
                b = d    
        return [a, b]

Comments

Post a Comment

Popular posts from this blog

  CA 09 - Move All Negative Elements to end Move all negative elements to end Difficulty:  Easy Accuracy:  56.24% Submissions:  174K+ Points:  2 Given an unsorted array  arr[ ]  having both negative and positive integers. The task is to place all negative elements at the end of the array without changing the order of positive elements and negative elements. Note :  Don't return any array, just in-place on the array. Examples: Input : arr[] = [1, -1, 3, 2, -7, -5, 11, 6 ] Output : [1, 3, 2, 11, 6, -1, -7, -5] Explanation: By doing operations we separated the integers without changing the order. Input : arr[] = [-5, 7, -3, -4, 9, 10, -1, 11] Output : [7, 9, 10, 11, -5, -3, -4, -1] MY CODE: class Solution: def segregateElements(self, arr): a = [] b = [] for i in arr: if i >= 0: a.append(i) else: b.append(i) temp = a + b fo...
Read more

CA 22 - Reverse a Linked List

Image
  Reverse a linked list Difficulty:  Easy Accuracy:  73.11% Submissions:  384K+ Points:  2 Average Time:  30m You are given the  head  of a singly  linked list . You have to  reverse  the linked list and return the  head  of the reversed list. Examples: Input: Output: 4 -> 3 -> 2 -> 1 Explanation: After reversing the linkedList Input: Output: 8 -> 9 -> 10 -> 7 -> 2 Explanation: After reversing the linked list Input: Output: 8 Explanation: Constraints: 1 ≤ number of nodes ≤ 10 5 1 ≤ node->data ≤ 10 5 MyCODE: class Solution:     def reverseList(self, head):         prev=None         curr=head         while curr:             n_x =curr.next             curr.next=prev             prev=curr     ...
Read more