Answer / senthilkumar

#include<iostream.h>
#include<conio.h>
void main()
{
clrscr();
int lsum,sum,i,j,n,a[20],flag=0;
lsum=0;sum=0;
cout<<"Enter the total no. of elements:";
cin>>n;

for(i=0;i<n;i++)
cin>>a[i];

//Checking condition whether there are only negative numbers
for(i=0;i<n;i++)
{
if(a[i]>0)
flag++;
}
//for array including positive and negative numbers
if(flag>0)
{
for(i=0;i<n;i++)
{
sum+=a[i];
if(sum<0)
{
sum=0;
j++;
i=j;
}
else if(lsum<sum)
lsum=sum;
}
}
//for array having only negative numbers
else
{
lsum=a[0];
for(i=0;i<n;i++)
{
if(lsum<a[i])
lsum=a[i];
}
}
cout<<"The largest sum is:"<<lsum;
getch();
}

Answer / raghuram.a

#include <iostream.h>
#include<conio.h>
/*prints subarray producing maximum sum..Efficiency is O(n)*/
main()
{
int flag=0,n,i,c=0,sum=0,max=0,a[25],s=-1,e=-1;
cout<<"enter n:";
cin>>n;
cout<<"enter elements:";
for(i=1;i<=n;i++)
cin>>a[i];
for(i=1;i<=n;i++)
{
if(a[i]+sum>0)
sum+=a[i];
else
{
sum=0;
if(c!=0)
c=1;
}
if(sum>max)
{
max=sum;
e=i;
s=e-c;
c++;
}
}
cout<<"max sum is:"<<max;
cout<<"\n\nmax sum producing sub array is:";
if(s!=-1&&e!=-1)
for(i=s;i<=e;i++)
cout<<"\t"<<a[i];
getch();
return 0;
}

Answer / saurabh

#include<stdio.h>
#include<stdlib.h>

int main()
{
int arr[50];
int i,sum1,sum2,n;
int flag;
//int front1,front2,rear1,rear2;
printf("How many elements do you want to enter: ");
scanf("%d",&n);
for(i=0;i<n;i++)
scanf("%d",&arr[i]);
sum1=arr[0];
sum2=arr[0];
if(arr[0]<0)
flag=1;
else
flag=0;
for(i=1;i<n;i++)
{
if(flag==1)
{

if((arr[i]>0&&arr[i-1]<0)||(arr[i]<0&&arr[i-1]<0&&arr[i]>arr[i-1]))
sum2=0;
if(arr[i]<0&&arr[i-1]<0&&arr[i]<arr[i-1])
{
int temp=arr[i];
arr[i]=arr[i-1];
arr[i-1]=arr[i];
sum2=0;
}
if(arr[i]<0&&arr[i-1]>0&&(arr[i-1]+arr[i])<0)
sum1=arr[i-1];
}
sum2=sum2+arr[i];
if(sum2>sum1)
{
sum1=sum2;
flag=0;
}
if(sum2<0)
{
sum2=0;
sum2=sum2+arr[i];
flag=1;
}
}
//printf("%d",rear);
if(sum1>=sum2)
printf("Largest subarray sum= %d\n",sum1);
else
printf("Largest subarray sum= %d\n",sum2);
return 0;
}

Answer / jjaspirin

I came to this website looking for the solution and finally
ended up writing it myself. Hope this is useful for someone
else.

Here is a working solution. Set the array "a" and N_ELEMENTS
accordingly. Some cases are not covered but should be an
easy fix.

#include <stdio.h>
#define N_ELEMENTS 7

int main() {
int a[N_ELEMENTS] = {-1, 2, -3, 2, 0, 5, -11 }; // if
you change the array, make sure you change N_ELEMENTS
int i = 0;

while(a[i] < 0 && i<N_ELEMENTS) {
i++;
}

if (a[i] < 0) {

printf ("DEBUG: array with only negative numbers.
Print the smallest negative number as the sum and we are
done.\n");

}

int sum_p=0, sum_n = 0;
int largest_sum = 0;

while (i<N_ELEMENTS) {
if (a[i] > 0) {
sum_p += a[i];

}
else {
sum_n += a[i];

}

if (sum_p+sum_n > largest_sum) {
largest_sum = sum_p + sum_n;

}

if (sum_p+sum_n <= 0) {
// find the next positive number
while(a[i] < 0 && i<N_ELEMENTS) {
i++;
}
if (a[i] < 0 || i == N_ELEMENTS) {
break;

}

sum_p = 0;
sum_n = 0;

} else {
i++;
}
}

printf ("DEBUG: The largest consecutive sum = %d\n",
largest_sum);

}

Answer / somebody

Guys..Answer 3 is correct..When there are -ve
numbers..maximum sum is 0.U should not consider any elements
at all..Coz null set which has maximum sum.i.e 0.So the 4th
ans is right.

Answer / raghuram.a

#include <iostream.h>
#include<conio.h>
main()
{
int s=-1,m=1,n,max=0,e=-1,sum=0,i,a[20];
cout<<"enter n:";
cin>>n;
cout<<"enter elements:";
for(i=0;i<n;i++)
cin>>a[i];
for(i=0;i<n;i++)
{
if(m==1)
{
if(a[i]>0)
s=i;
m=0;
}
if(a[i]+sum>0)
sum+=a[i];
else
{

m=1;
sum=0;
}
if(sum>max)
{
max=sum;
e=i;
}
}
cout<<"max sum is:"<<max;
cout<<"\n\nmax sum producing sub array is:";
if(s!=-1&&e!=-1)
for(i=s;i<=e;i++)
cout<<a[i]<<"\t";
getch();
return 0;
}

Answer / dilip

i think ths program doesnt not work when the sum is
negative.. cant we not work around it by setting the max as
somelarge negative value and comparing with that??

Answer / deepika jadav

the cod is written in c#...similar to java and c.....so interpret it accordingly..

static void Main(string[] args)
{
//int[] arr={2,3,-5,-2,-2,5,6,7,8,-2,3,4};
int[] arr = {-2,-3,-4,-5,-6,-7 };
int max=0;
int sum=0;

int min = 0;
int flag = 0;
for (int i = 0; i < arr.Length;i++)
{
if (arr[i] < 0)
{
sum = 0;
flag++;
if(flag==1)
min = arr[i];
if (flag == arr.Length)
Console.WriteLine("The smallest no amogst negative nos is {0} hence the largest aum is {1}",min,min);
if (min < arr[i])
min = arr[i];
continue;
}
else
{
sum = sum + arr[i];
}

if (max < sum)
max = sum;


}
if(flag!=arr.Length)
Console.WriteLine("the Largest Sum is {0}",max);
Console.ReadLine();
}

Answer / raghuram.a

Actually I have posted 3 answers..Bcoz after posting twice I
found there are some logical errors.Last one is fully
correct..If not please let me know..

Answer / raghuram.a

/*prints subarray producing maximum sum.Efficiency is O(n)*/
#include <iostream.h>
#include<conio.h>
main()
{
int n,i,c=0,sum=0,max=0,a[25],s=-1,e=-1;
cout<<"enter n:";
cin>>n;
cout<<"enter elements:";
for(i=1;i<=n;i++)
cin>>a[i];
for(i=1;i<=n;i++)
{
if(a[i]+sum>0)
{
sum+=a[i];
if(sum>=max)
s=i-c;
c++;
}
else
{
sum=0;
c=0;
}
if(sum>max)
{
max=sum;
e=i;
}
}
cout<<"max sum is:"<<max;
cout<<"\n\nmax sum producing sub array is:";
if(s!=-1&&e!=-1)
for(i=s;i<=e;i++)
cout<<"\t"<<a[i];
getch();
return 0;
}

main() { int c[ ]={2.8,3.4,4,6.7,5}; int j,*p=c,*q=c; for(j=0;j<5;j++) { printf(" %d ",*c); ++q; } for(j=0;j<5;j++){ printf(" %d ",*p); ++p; } }

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