Array processing (elimination of three largest values) (one of many array reduction problems) The array a(1..n) contains arbitrary integers. Write a function reduce(a, n) that reduces the array a(1..n) by eliminating from it all values that are equal to three largest different integers. For example, if a=(9,1,1,6,7,1,2,3,3,5,6,6,6,6,7,9) then three largest different integers are 6,7,9 and after reduction the reduced array will be a=(1,1,1,2,3,3,5), n=7. The solution should have the time complexity O(n).

Step-by-step explanation:

# include <bits/stdc++.h>

using namespace std;

bool find3Numbers(int A[], int arr_size, int sum)

{

int l, r;

sort(A, A+arr_size);

for (int i=0; i<arr_size-2; i++)

{

l = i + 1;

r = arr_size-1;

while (l < r)

{

if( A[i] + A[l] + A[r] == sum)

{

printf("Triplet is %d, %d, %d", A[i],

A[l], A[r]);

return true;

}

else if (A[i] + A[l] + A[r] < sum)

l++;

else // A[i] + A[l] + A[r] > sum

r--;

}

}

return false;

}

int main()

{

int A[] = {1, 4, 3, 2, 10, 8};

int sum = 22;

int arr_size = sizeof(A)/sizeof(A[0]);

find3Numbers(A, arr_size, sum);

return 0;

}

a

step-by-step explanation:

when a number is between two straight lines it means the absolute value.

to find absolute value you need to know how many numbers is that number away from 0.

when finding absolute value the number will always come out positive

so absolute value of -3 is 3

absolute value of 3/2 is 3/2

absolute value of -2 is 2

absolute value of 3.5 is 3.5

absolute value of -1 is 1

now put these numbers in order:

a)   |-1|, |3/2|, |-2|, |-3|, |3.5|