Given an array whose elements represents the coefficients of a polynomial of degree n, if the polynomial has a degree n then the array will have n+1 elements (one extra for the constant of a polynomial). Swap some elements of the array and print the resulting array such that the sum of the roots of the given polynomial is as minimum as possible irrespective of the nature of the roots.
Note that : except the first element of the array elements can be 0 also and the degree of the polynomial is always greater than 1.
Input : -4 1 6 -3 -2 -1 Output : 1 6 -4 -3 -2 -1 Here, the array is -4, 1, 6, -3, -2, -1 i.e the polynomial is -4.x^5 + 1.x^4 + 6.x^3 - 3.x^2 - 2.x^1 - 1 minimum sum = -6 Input : -9 0 9 Output :-9 0 9 Here polynomial is -9.x^2 + 0.x^1 + 9 minimum sum = 0
Solution : Let us recall the fact about the sum of the roots of a polynomial if a polynomial p(x) = a.x^n + b.x^n-1 + c.x^n-2 + … + k, then the sum of roots of a polynomial is given by -b/a. Please see Vieta’s formulas for details.
We have to minimize -b/a i.e to maximize b/a i.e maximize b and minimize a. So if somehow we are able to maximize b and minimize a, we will swap the values of the coefficients and copy rest of the array as it is.
There will be four cases :
Case #1: when the number of positive coefficients and the number of negative coefficients both are greater than or equal to 2
In this case, we will find a maximum and minimum from positive elements and from negative elements also and we will check -(maxPos)/(minPos) is smaller or -( abs(maxNeg) )/ ( abs(minNeg) ) is smaller and print the answer after swapping accordingly.
Case #2: when the number of positive coefficients is greater than equal to 2 but the number of negative coefficients is less than 2
In this case, we will consider the case of the maximum of positive and minimum of positive elements only. Because if we picked up one from positive elements and the other from negative elements the result of -b/a will be a positive value which is not minimum. (as we require a large negative value)
Case #3: when the number of negative coefficients is greater than equal to 2 but the number of positive coefficients is less than 2
In this case, we will consider the case of the maximum of negative and minimum of negative elements only. Because if we picked up one from positive elements and the other from negative elements the result of -b/a will be a positive value which is not minimum. (as we require a large negative value)
Case #4: When both the counts are less than or equal to 1
Observe carefully, You cannot swap elements in this case.
1 6 -4 -3 -2 -1
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