There is an array A[N] of N numbers. You have to compose an array Output[N] such that Output[i] will be equal to multiplication of all the elements of A[N] except A[i]. For example Output[0] will be multiplication of A[1] to A[N-1] and Output[1] will be multiplication of A[0] and from A[2] to A[N-1].

3 thoughts on “There is an array A[N] of N numbers. You have to compose an array Output[N] such that Output[i] will be equal to multiplication of all the elements of A[N] except A[i]. For example Output[0] will be multiplication of A[1] to A[N-1] and Output[1] will be multiplication of A[0] and from A[2] to A[N-1].”

1. Pa1

   17 years ago

   With Division operator, this would be a piece of cake…

   Pa1

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2. Esu

   14 years ago

   Easy, one can use that a = exp(ln(a)) or a = 10^log(a).
   This pseudocode solution needs all A[i] to be nonzero:

   TMP := 0
   SN := 1
   FOR i=0 TO N-1
   TMP := TMP + ln( abs( A[i] ) )
   SN := SN*sgn( A[i] )
   END

   FOR i=0 TO N-1
   OUT[i] := TMP – ln( abs( A[i] ) )
   OUT[i] := OUT[i]*SN*sgn( A[i] )
   END

   Where “ln” is the natural logarithm, “abs” is the absolute value and “sgn” is the sign.
   This should work, though I didn’t test it.

   Greetings, Esu

   Reply
3. Pa1

   14 years ago

   @Esu,

   What if one of the array element is zero ?

   Also, what is the complexity of ur algo ?

   – Pa1

   Reply