Today we are going to discuss some important Accenture Coding Questions and answers that has been asked in previous years for engineering Fresher hiring.
So, If you are also appearing in Accenture, then there are fair chances that you can get same difficulty level of questions in your actual exam.
Electrostatic Field
Doug is fond of change, every now and then he tries to do new things. This time, he caught up with a rod comprising of negative (N) and positive (P) charges. He is asked to calculate the maximum net electrostatic field possible in the region due to the rod.
Note: Assume Electrostatic Field = Total charge * 100
Input specification:
Input 1: Integer array denoting the magnitude of each charge.
Input 2: String denoting nature of each charge ith represents a sign of charge at ithentry represents a sign of charge at ith location in input1
Input 3: No of charges it holds (length of input1)
Output Specification:
Return the next maximum electrostatic field possible in the rod.
Example 1:
Input 1: {4,3,5}
Input 2: PNP
Input 3: 3
Output: 600
Explanation:
The maximum electric charge on the rod is 4-3+5 = 6 units. So the magnitude of the electric field would be 6*100=600
Example 2:
Input 1: {2,3}
Input 2: PN
Input 3: 2
Output : 100
Explanation:
The maximum possible electric charge on the section of the rod is 2-3=-1 unit.
Solution in C :
#include <stdio.h>
#include <string.h>
int main()
{
char a[100];
gets(a);
char s[100];
gets(s);
int n;
scanf("%d",&n);
int charge[n];
int count = 0;
for(int i=0;i<strlen(a);i++){
if(a[i] != ' ') {
charge[count] = a[i]-48;
count++;
}
}
int sum = 0;
for(int i=0;i<n;i++) {
if(s[i] == 'P')
sum += charge[i];
else
sum -= charge[i];
}
if(sum > 0) {
printf("%d",sum*100);
}
else {
printf("%d",-sum*100);
}
}Output:
4 3 5
PNP
3
600