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#include
#include <stdio.h>
#include
int main() {
int cases, i;
scanf("%d", &cases);
for (i=0; i<cases; i++) {
int n,j,k;
long int a[100000];
scanf("%d", &n);
for (j=0; j<n; j++)
std::cin>>a[j];
for (j=0; j<n; j++) {
int count=1;
for (k=j; k<n; k++) {
//printf("%d and %d\n", a[k], a[k+1]);
if (a[k]*a[k+1] < 0)
count++;
else
break;
}
printf("%d ", count);
}
printf("\n");
}
return 0;
}
What is wrong in this code, Why is it showing Wrong Answer !
All the test cases turn out fine.
#include
using namespace std;
int main()
{
int t;
cin>>t;
while(t–)
{
int n;
cin>>n;
string s;
long long int x;
int a[n];
for(int i=0;i<n;i++)
{
cin>>x;
if(x>0)
s+=’+’;
else
s+=’-’;
a[i]=1;
}
//cout<<s<<endl;
for(int i=1;i<n;i++)
{
if(s[i]!=s[i-1])
a[0]++;
else
break;
}
cout<<a[0]<<" “;
for(int i=1;i<n;i++)
{
if(a[i-1]>1)
{
a[i]=a[i-1]-1;
}
else
for(int j=i+1;j<n;j++)
{
if(s[j]!=s[j-1])
a[i]++;
else
break;
}
cout<<a[i]<<” ";
}
cout<<endl;
}
return 0;
}
I am wondering if you could solve that problem using a segment tree ? Anyone used that approach ? Would it be more efficient ?
Thanks
One question we can ask in an interview will be to how to treat 0’s .
#include<stdio.h>
#define MAX 1000007
typedef long long int ll;
int stk[MAX];
int top = -1;
int empty(){
return top == -1;
}
void push(int x){
stk[++top] = x;
}
int pop(){
if(!empty())
return stk[top--];
}
int peek(){
if(!empty())
return stk[top];
}
int main(int argc, char const *argv[])
{
/* code */
int t;
scanf("%d",&t);
while(t--){
int n;
scanf("%d",&n);
int res[n],s[n];
ll a[n];
for(int i=0;i<n;i++){
scanf("%lld",a+i);
res[i] = 1;
s[i] = (a[i]>0)? 1 : -1;
}
top=-1;
int cnt=0;
for(int i=0;i<n;i++){
if(!empty() && (s[peek()] == s[i]) ){
cnt=0;
while(!empty()){
res[peek()] += cnt++;
pop();
}
}
else
push(i);
}
cnt=0;
while(!empty()){
res[peek()] += cnt++;
pop();
}
for(int i=0;i<n;i++)
printf("%d ",res[i]);
printf("\n");
}
return 0;
}
I want to know on which test case this stack implementation fails…help…
def getAlt(arr):
n = len(arr)
ans = [1]*n
for i in range(n-2, -1, -1):
# print(ans[i], ans[i+1])
if (arr[i] < 0 and arr[i+1] > 0) or (arr[i] > 0 and arr[i+1] < 0):
ans[i] += ans[i+1]
# print(ans)
print(’ '.join(str(x) for x in ans))
for _ in range(int(input())):
n = int(input())
arr = list(map(int, input().split()))
getAlt(arr)
A simple pytonh DP approach utilising the fact that minimum length will be taken as 1 per index
Why can’t we do this?
- Change every positive element to 1 and every negative element to -1
- Then calculate to product of every two consective terms of the new array
- Check the maximum number of continuous -1 's in the product array
- The answer is max number of -1 's (-1)