Question

Write an algorithm to find the sum of energy
of the two chemicals which produces
maximum energy on reaction.
Input
The first line of the input consists of an integer
- numOfChem, representing the number of
chemicals (N).
The second line consists of N space-separated
integers - enery, enerz....., energrepresenting
the energies of the chemicals.
Output
Print an integer representing the sum of
energy of the two chemicals which produces
maximum energy on reaction.
Constraints
os numorChem < 106
-10€ s ener, s 10%
1 $i$ numOfChem​

Answer 1

Answer:

Given an array containing positive and negative numbers. The array represents checkpoints from one end to other end of street. Positive and negative values represent amount of energy at that checkpoint. Positive numbers increase the energy and negative numbers decrease. Find the minimum initial energy required to cross the street such that Energy level never becomes 0 or less than 0.

Note : The value of minimum initial energy required will be 1 even if we cross street successfully without loosing energy to less than and equal to 0 at any checkpoint. The 1 is required for initial check point.

Examples :

Input : arr[] = {4, -10, 4, 4, 4}

Output: 7

Suppose initially we have energy = 0, now at 1st

checkpoint, we get 4. At 2nd checkpoint, energy gets

reduced by -10 so we have 4 + (-10) = -6 but at any

checkpoint value of energy can not less than equals

to 0. So initial energy must be at least 7 because

having 7 as initial energy value at 1st checkpoint

our energy will be = 7+4 = 11 and then we can cross

2nd checkpoint successfully. Now after 2nd checkpoint,

all checkpoint have positive value so we can cross

street successfully with 7 initial energy.

Input : arr[] = {3, 5, 2, 6, 1}

Output: 1

We need at least 1 initial energy to reach first

checkpoint

Input : arr[] = {-1, -5, -9}

Output: 16

Recommended: Please solve it on “PRACTICE” first, before moving on to the solution.

We take initial minimum energy 0 i.e; initMinEnergy = 0 and energy at any checkpoint as currEnergy = 0. Now traverse each checkpoint linearly and add energy level at each i’th checkpoint i.e; currEnergy = currEnergy + arr[i]. If currEnergy becomes non-positive, then we need at least “abs(currEnergy) + 1” extra initial energy to cross this point. Therefore we update initMinEnergy = (initMinEnergy + abs(currEnergy) + 1). We also update currEnergy = 1 as we now have the required extra minimum initial energy for next point.

Answer 2

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