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50 lines
1.3 KiB
Ruby
50 lines
1.3 KiB
Ruby
#Given an unsorted array,Find a pair of elemens with given difference,if no such pair exists return -1.
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#Approach 1: Sort the array then search using two pointers left and right
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#Time-complexity: O(nlogn),Auxiliary-space:O(1)
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def check_pair(a,x)
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len=a.length
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a.sort!
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left=0
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right=len-1
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while left<right
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diff=a[right]-a[left]
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if diff==x.abs #Considering absolute difference as pair a,b and b,a have same absolute difference
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return a[left],a[right]
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elsif diff>x.abs
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right-=1
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else
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left+=1
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end
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end
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return -1
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end
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check_pair([5, 20, 3, 2, 50, 80],-78) # => [2, 80]
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check_pair([5, 20, 3, 2, 50, 80],78) # => [2, 80]
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check_pair([90, 70, 20, 80, 50],45) # => -1
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#Approach 2: Using hash maps
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#Time-complexity: O(n), Auxiliary-space: O(n){for hash}
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def check_pair(a,x)
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len=a.length
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map=Hash.new()
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for i in 0...len
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t=x.abs+a[i] #if a-b=x then a=x+b
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s=a[i]-x.abs #if a-b=x then b=x-a
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if map[t]
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return t,a[i]
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elsif map[s]
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return s,a[i]
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end
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map[a[i]]=1
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end
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return -1
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end
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check_pair([5, 20, 3, 2, 50, 80],-78) # => [2, 80]
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check_pair([5, 20, 3, 2, 50, 80],78) # => [2, 80]
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check_pair([90, 70, 20, 80, 50],45) # => -1
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