124 lines
3.1 KiB
Ruby
124 lines
3.1 KiB
Ruby
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# Given an array of n+2 elements with elements value lying beween 1 to n.
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# All elements occur once except two.Print the two repeating numbers.
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#Approach 1: Using sum and product of array (Making two equation)
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#Time-complexity: O(n) Auxiliary-space: O(1)
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#Caveat: If array is large this can lead to integer overflow.
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def find_duplicates(a)
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len=a.length
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n=len-2
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sum_of_series= (n*(n+1))/2 #Sum of series 1 to n
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#Find sum and product of array elements
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sum=0
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product=1
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for x in a
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sum+=x
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product*=x
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end
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sum-=sum_of_series #sum is a+b now,where a and b are repeating numbers
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product/=fact(n) #product is a*b now,where a and b are repeating numbers
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temp= Math.sqrt((sum**2)-(4*product)).to_i # temp is a-b
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print "#{(sum+d)/2} #{(sum-d)/2}"
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end
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def fact(x)
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return 1 if (x==0 || x==1)
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factorial=1
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while x>1
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factorial*=x
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x-=1
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end
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return factorial
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end
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find_duplicates([1,2,3,4,5,5,6,7,8,8]) # => 5 8
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#Approach 2: Use count array
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#Time-complexity: O(n) Auxiliary-space: O(n) {for count array}
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def find_duplicates(a)
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len=a.length
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n=len-2
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count=Array.new(n,0) # count has length n
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for i in 0...len
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if count[a[i]-1]==1
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print "#{a[i]} "
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else
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count[a[i]-1]+=1
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end
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end
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return
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end
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find_duplicates([1,2,3,4,5,5,6,7,8,8]) # => 5 8
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#Approach 3: Using Array Indexing
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#Time-complexity: O(n) Auxiliary-space: O(1)
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#Caveat: This algorithm modifies the array-elements
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def find_duplicates(a)
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len=a.length
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for i in 0...len
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if a[a[i].abs]<0
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print "#{a[i].abs} "
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else
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a[a[i].abs]= -a[a[i].abs]
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end
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end
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return
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end
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find_duplicates([1,2,3,4,5,5,6,7,8,8]) # => 5 8
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#Approach 4: Using bitwise XOR,take xor of all elements and all numbers from 1 to n then,
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#find rightmost set bit,divide the range in two sets and take xor of each set, first with xor then with range.
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#Time-complexity: O(n) Auxiliary-space: O(1)
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def find_duplicates(a)
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len=a.length
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n=len-2
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xor= 0
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x,y=0,0 #variables to store duplicates
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#xor of all numbers from 1 to n
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for i in 1..n
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xor^=i
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end
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#xor of all array elements
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for i in 0...len
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xor^=a[i]
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end
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#Rightmost set bit
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set_bit_pos= xor & ~(xor-1)
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#Divinding array in two sets ,one with set bit at set_bit_pos and other with 0.
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for i in 0...len
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if (a[i] & set_bit_pos == 0)
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x^=a[i] # XOR of first-set(with 0 at set_bit_pos) in array
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else
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y^=a[i] # XOR of second-set(with 1 at set_bit_pos) in array
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end
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end
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for i in 0..n
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if (i & set_bit_pos == 0)
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x^=i # XOR of first-set(with 0 at set_bit_pos) in range
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else
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y^=i # XOR of second-set(with 1 at set_bit_pos) in range
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end
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end
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print "#{x} #{y}"
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return
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end
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find_duplicates([1,2,3,4,5,5,6,7,8,8]) # => 5 8
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