Answer is 15
How?
Its simple, just multiply the number by 3.
How?
Its simple, just multiply the number by 3.
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Explanation:
1 + 2 => (1)^2 + (1*2) = 3
3 + 4 => (3)^2 + (3*4) = 21
5 + 6 => (5)^2 + (5*6) = 55
Thus 4 + 5 => (4)^2 + (4*5) = 36
Problem Statement:
Given a set S of n integers find all possible subsets(a,b,c) such that a + b + c = 0.
Solution:
Brute force approach is of O(n^4) but we can solve it in O(n^2) by using the approach in Non duplicate pairs that sum to S.
First sort the array(Order O(nlogn)), than finding a, b, c pairs is equal to finding=> For every element a in the array, if there exists a pair with sum equal to -a. As explained in Non duplicate pairs that sum to S, we can get the pair with sum -a in O(n) and we have to repeat this exercise n times so order of complexity will be O(n^2).
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How?
If you see the rows every next number = previous number in the row +3.
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How?
Number on the bigger line is the multiplication of the numbers on the adjacent smaller lines, thus the answer should be 10*9=90.
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How?
Actually number on the right side is the square of the exact opposite number.
thus the missing number should be 6^2.
How?
There is gap of 4 between numbers
8+4 = 12
3+4 = 7
9-5 = 4
simillarly
?-4 = 6=> ?=10
NOTE: there is a better and shorter explanation(suggested by Guestlogic in comment), opposite numbers sum is 15, that’s why answer should be 15 – 5 = 10
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