Hackerrank's Problem Solving solutions( A Very Big Sum, Diagonal Difference, Plus Minus ) :

Problem 4: A Very Big Sum

Solution: (in c++)

( please guys before moving to the solution try it yourself at least 3-4 times , if you really wanna become a good coder)

#include<bits/stdc++.h>

using namespace std;

int main()

{

int n,j,i;

long long sum;

cin>>n;

int a[n];

for(i=0;i<n;i++)

cin>>a[i];

sum=0;

for(j=0;j<n;j++)

{

sum=sum+a[j];

}

cout<<sum;

}

This code is simple. There's no need for any explanation.
 

 

 

 

Problem 5: Diagonal Difference

Solution: (in c++)

( please guys before moving to the solution try it yourself at least 3-4 times , if you really wanna become a good coder)

#include<bits/stdc++.h>

using namespace std;

int main()

{

int n,a[100][100],sum,summ,i,d,j; //summ = sum of '/' diagonal

cin>>n;

for(i=0;i<n;i++)

for(j=0;j<n;j++)

{

cin>>a[i][j]; //made a matrix

}

sum=0;

for(i=0;i<n;i++)

{

sum=sum+a[i][i]; //sum = sum of '\' diagonal

}

summ=0;

j=n-1;

for(i=0;i<n;i++)

{

summ=summ+a[i][j]; //see note below

j--;

}

d=abs(sum-summ);

cout<<d;

}

Let a 3*3 matrix and the position of  a given element is represented in the form of (i,j) :

       0       1        2
0  (0,0)  (0,1)  (0,2)

1  (1,0)  (1,1)  (1,2)

2  (2,0)  (2,1)  (2,2)

Here we can see  for diagonal \ , the elements can be represented by (i,i) but for the diagonal / , the elements can't be represented in a general form. But we can observe that on traversing this diagonal from top right to bottom left i value increases by 1 while j value decreases by 1. This is what i have done in the above code.

 

 

Problem 6: Plus Minus

Solution: (in c++)

( please guys before moving to the solution try it yourself at least 3-4 times , if you really wanna become a good coder)

#include<bits/stdc++.h>

using namespace std;

int main()

{

int a[100];

int n;

float c1=0,c2=0,c3=0;

cin>>n;

for(int i=0;i<n;i++)

{

cin>>a[i];

}

for(int j=0;j<n;j++)

{

if(a[j]>0)

{

c1++;

}

else if(a[j]==0)

{

c2++;

}

else {

c3++;

}

}

float r1,r2,r3;

r1=c1/n;

r2=c2/n;

r3=c3/n;

cout<<r1<<endl<<r3<<endl<<r2;

}

This code is simple. There's no need for any explanation.
 

 

 

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