Find two repeating elements

Author: kumar91gopi

arrays/FindTwoRepeatingElem.rb

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