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Python – Rearrange elements second index greater than first

  • Last Updated : 10 Jul, 2020
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Given 2 lists, for a given index, 2nd list element is always larger than first, and if not, we rearrange it.

Input : test_list1 = [36, 38, 40, 132], test_list2 = [35, 37, 39, 41, 133]
Output : [37, 39, 41, 133]
Explanation : Each element in result list is greater than its index counterpart of 1st list. (Eg. 37 > 36)

Input : test_list1 = [2, 6], test_list2 = [5, 3, 8]
Output : [5, 8]
Explanation : Here 5 > 2 and 8 > 6.

Method : Using loop
This is brute way to tackle this problem. In this, we try to get best suitable next higher element after the whole list traversal and perform the necessary rearrangement.




# Python3 code to demonstrate working of 
# Rearrange elements second index greater than first
# Using loop
  
# initializing lists
test_list1 = [14, 16, 18, 110]
test_list2 = [13, 15, 17, 19, 111]
  
  
# printing original lists
print("The original list 1 is : " + str(test_list1))
print("The original list 2 is : " + str(test_list2))
  
# Rearrange elements second index greater than first
# Using loop
x = y = 0
res1, res2 = [], []
while x < len(test_list2) and y < len(test_list1):
      
    # checking for greater element
    if test_list2[x] > test_list1[y]:
        res2.append(test_list2[x])
        res1.append(test_list1[y])
        while y < len(test_list1) and test_list2[x] > test_list1[y]:
            res1[-1] = test_list1[y]
            y += 1
    x += 1
  
# printing result 
print("List 2 after conversion : " + str(res2))
Output :



The original list 1 is : [14, 16, 18, 110]
The original list 2 is : [13, 15, 17, 19, 111]
List 2 after conversion : [15, 17, 19, 111]

 

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