Python – Find all pairs of consecutive odd positive integer smaller than a with sum greater than b
Last Updated :
04 Apr, 2023
Given two positive integers a and b, the task is to write a program in python to find all pairs of consecutive odd numbers which are smaller than the first number a and their sum should be greater than the second number b.
Examples:
Input:
a = 60
b = 100
Output:
Pairs of consecutive number are:
51 , 53
53 , 55
55 , 57
57 , 59
Input:
a = 20
b = 200
Output:
None
Approach:
Two numbers are given and then check if they are a positive integer and checked for the first number to be greater than the half of the second number. Then we will check for odd positive integer and assigned in a variable a. In the while statement, pairs of the odd consecutive integers are found and printed.
Example 1:
Python3
a = 60
b = 100
print('a =', a)
print('b =', b)
if(a > 0 and b > 0 and a > b/2):
if(a % 2 == 0):
a -= 1
else:
a -= 2
b //= 2
print("Pairs of consecutive number are:")
while(b <= a):
if(b % 2 != 0):
x = b
if(x + 2 <= a):
print(x, ',', x+2)
b += 1
else:
print("None")
|
Output:
a = 60
b = 100
Pairs of consecutive number are:
51 , 53
53 , 55
55 , 57
57 , 59
Time complexity: O(n)
Auxiliary space: O(1)
Example 2:
Python3
a = 20
b = 200
print('a =', a)
print('b =', b)
if(a > 0 and b > 0 and a > b/2):
if(a % 2 == 0):
a -= 1
else:
a -= 2
b //= 2
print("Pairs of consecutive number are:")
while(b <= a):
if(b % 2 != 0):
x = b
if(x + 2 <= a):
print(x, ',', x+2)
b += 1
else:
print("None")
|
Output:
a = 20
b = 200
None
Complexity analysis for all examples :
Time complexity: O(n) where n is the length of the string.
Auxiliary Space: O(1)
Iterative Search for Consecutive Odd Integer Pairs
Steps:
- Iterate through all odd integers from 1 to a-1.
- For each odd integer i, check if the next odd integer i+2 is also less than a.
- If i and i+2 are both less than a, check if their sum is greater than b.
- If the sum of i and i+2 is greater than b, add the pair (i, i+2) to a list of pairs.
- Return the list of pairs.
Python3
def find_pairs(a, b):
pairs = []
for i in range(1, a-1, 2):
if i+2 < a:
if i + i+2 > b:
pairs.append((i, i+2))
return pairs
a = 60
b = 100
pairs = find_pairs(a, b)
print("Pairs of consecutive number are:")
for pair in pairs:
print(pair[0], ",", pair[1])
|
Output
Pairs of consecutive number are:
51 , 53
53 , 55
55 , 57
57 , 59
The time complexity of this approach is O(a).
The auxiliary space required is O(1)
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