Install Social Engineer Toolkit in termux 2022 Without errors

Hi friends, we are going to solve a leetcode problem called Merge Sorted Array.  Iam to to write a series of posts regarding solving the problem on leet code on wards. Let's check the problem 

Description of the problem:

You are given two integer arrays nums1 and nums2, sorted in non-decreasing order, and two integers m and n, representing the number of elements in nums1 and nums2 respectively.

Our task is to merge nums1 and nums2 into a single array sorted in the non-decreasing order. The final sorted array should not be returned by the function but instead be stored inside the array nums1. To accommodate this, nums1 has a length of m + n, where the first m elements denote the elements that should be merged, and the last n elements are set to 0 and should be ignored. nums2 has a length of n.

Example 

Input: 

nums1 = [1,2,3,0,0,0], m = 3

nums2 = [2,5,6], n = 3

Output: 

[1,2,2,3,5,6]

Approach

To solve this problem efficiently, we can use a two-pointer technique. This approach allows us to avoid the need for extra space and directly merge the arrays in place.

Steps:

1. Initialise pointers
• Initialize Pointers:
• One pointer (i) for traversing nums1 starting from the end of its valid elements.
• Another pointer (j) for traversing nums2 from the end.
• A third pointer (last) for placing elements in their correct positions starting from the end of nums1.
• Declaration of pointers

last=m+n-1

i=m-1

j=n-1

2. Merge in Reverse Order:
   • Compare elements pointed by i and j.
   • Place the larger element at the position pointed by last.
   • Move the respective pointers (i, j, last) accordingly.
   • Continue this until all elements from nums2 are placed in nums1

Python

def merge(self, nums1, m, nums2, n):
    last=m+n-1
    i=m-1
    j=n-1
    while j>=0:
        if i>=0 and nums1[i]>nums2[j]:
            nums1[last]=nums1[i]
            i-=1
        else:
            nums1[last]=nums2[j]
            j-=1