CODEWITHSUNDEEP
c++algorithm
Luv
Luv

Reputation: 5451

To find the min and max after addition and subtraction from a range of numbers

I am having a Algorithm question, in which numbers are been given from 1 to N and a number of operations are to be performed and then min/max has to be found among them.

Two operations - Addition and subtraction

and operations are in the form a b c d , where a is the operation to be performed,b is the starting number and c is the ending number and d is the number to be added/subtracted

for example

suppose numbers are 1 to N and N =5

1 2 3 4 5

We perform operations as

1 2 4 5

2 1 3 4

1 4 5 6

By these operations we will have numbers from 1 to N as

1 7 8 9 5

-3 3 4 9 5

-3 3 4 15 11

So the maximum is 15 and min is -3

My Approach: I have taken the lower limit and upper limit of the numbers in this case it is 1 and 5 only stored in an array and applied the operations, and then had found the minimum and maximum.

Could there be any better approach?

Upvotes: 0

Views: 3693

Answers (2)

nhahtdh
nhahtdh

Reputation: 56809

I will assume that all update (addition/subtraction) operations happen before finding max/min. I don't have a good solution for update and min/max operations mixing together.

You can use a plain array, where the value at index i of the array is the difference between the index i and index (i - 1) of the original array. This makes the sum from index 0 to index i of our array to be the value at index i of the original array.

Subtraction is addition with the negated number, so they can be treated similarly. When we need to add k to the original array from index i to index j, we will add k to index i of our array, and subtract k to index (j + 1) of our array. This takes O(1) time per update.

You can find the min/max of the original array by accumulating summing the values and record the max/min values. This takes O(n) time per operation. I assume this is done once for the whole array.

Pseudocode:

a[N] // Original array
d[N] // Difference array

// Initialization
d[0] = a[0]
for (i = 1 to N-1)
    d[i] = a[i] - a[i - 1]

// Addition (subtraction is similar)
add(from_idx, to_idx, amount) {
    d[from_idx] += amount
    d[to_idx + 1] -= amount
}

// Find max/min for the WHOLE array after add/subtract
current = max = min = d[0];
for (i = 1 to N - 1) {
   current += d[i]; // Sum from d[0] to d[i] is a[i]
   max = MAX(max, current);
   min = MIN(min, current);
}

Upvotes: 2

Quick n Dirty
Quick n Dirty

Reputation: 569

Generally there is no "best way" to find the min/max in the performance point of view because it depends on how this application will be used.

-Finding the max and min in a list needs O(n) Time, so if you want to run many (many in the context of the input) operations, your approach to find the min/max after all the operations took place is fine.

-But if the list will hold many elements and you don’t want to run that many operations, you better check each result of the op if its a new max/min and update if necessary.

Upvotes: 1

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