Swapping Pairs Make Sum Equal || Program 56 || Competitive Coding || Learning Monkey ||
Swapping Pairs Make Sum Equal
In this class, We discuss Swapping Pairs Make Sum Equal.
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Question:
Given two arrays of size N and M.
Our task is to find a pair of elements from each array.
Swapping the elements makes the sum of elements in both arrays is same.
Example:
N = 6, M = 4
A = [4, 1, 2, 1, 1, 2]
B = [3, 6, 3, 3]
Output = 1
Swap elements one from A and three from B.
After swapping, the sum of elements is the same in both arrays.
There is a possibility to swap, so output 1.
If there is no such possibility, Then output -1.
Logic:
Two ways we can solve this example.
1) Using sorting
2) Using Heap
Both ways are important to understand.
The step-by-step explanation is provided in the video.
Code:
class Solution:
def findSwapValues(self,a, n, b, m):
sum1,sum2=0,0
for i in range(n):
sum1+=a[i]
for j in range(m):
sum2+=b[j]
a.sort()
b.sort()
i,j=0,0
while(iltn and jltm):
val=sum1-(a[i])+(b[j])
val2=sum2+(a[i])-(b[j])
if val==val2:
return 1
if(valgtval2):
i+=1
else:
j+=1
return -1
a=[4, 1, 2, 1, 1, 2]
b = [3, 6, 3, 3]
ob=Solution()
z=ob.findSwapValues(a,6,b,4)
print(z)
class Solution:
def findSwapValues(self,a, n, b, m):
x, y = {}, {}
s1, s2 = 0, 0
for i in range(n):
s1 += a[i]
x[a[i]] = x.get(a[i], 0) + 1
for i in range(m):
s2 += b[i]
y[b[i]] = y.get(b[i], 0) + 1
if (s1 - s2) % 2:
return -1
for p in x:
q = ((s2 - s1)//2) + p
if q in y:
return 1
return -1
a=[4, 1, 2, 1, 1, 2]
b = [3, 6, 3, 3]
ob=Solution()
z=ob.findSwapValues(a,6,b,4)
print(z)
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