CareerCup

Recurssion can be used,

Int getLongestPalindromeLength(char inputString[], int start, int end)
{
If(start == stop) return 1;
Int i = start, j = stop;
for(; i< j; i++ ){
If(inputString[i] != inputString[j])
Break;
i++; j--;
}
If(i == j) return (Start - stop + 1);
getLongestPalindromeLength(inputString[], start-1, stop);
getLongestPalindromeLength(inputString[], start, stop -1);
getLongestPalindromeLength(inputString[], start-1, stop -1);
}

Recurssion can be used,

Int getLongestPalindromeLength(char inputString[], int start, int end)
{
If(start == stop) return 1;
Int i = start, j = stop;
for(; i< j; i++ ){
If(inputString[i] != inputString[j])
Break;
i++; j--;
}
If(i == j) return (Start - stop + 1);
getLongestPalindromeLength(inputString[], start-1, stop);
getLongestPalindromeLength(inputString[], start, stop -1);
getLongestPalindromeLength(inputString[], start-1, stop -1);
}

Critical section is a segment of code which can be executed by one process at a time and other processes should wait at that time. Example is writing database or file, which are read by other processes or using a shared resourse.

MUTEX is a variable used as a token to execute the critical section.

I am having few questions as above solution is wrong,

1. We should consider start and end point, as given in question.
2. Is travelling from any point to any point is posibble?

@decoder: can you plz explain the code..with significance of d used.

We cant store duplicate data, as it will lead to waste lot of memory. Instead we can have arrays, one for each key.
1. FirstName[0..length], lastName[0..length], number[0..length], designation[0..length]
2. Search value in respective array, which return the index of item.
3. Fetch data from other arrays from same index. This will return complete details of person.

Why not use a simple array..following is pseudocode..

1. Sort array A[0..length-1] and get sum of all elements as totalSum.
2. Set start counter to 0 and end counter at lenght - 1.
3. Get sum currentSum = A[start] + A[end].
4. Get sum restElementSum = totalSum - currentSum.
5. If restElementSum == currentSum then print A[start] and A[end] return
6. Else start-- and end--
7. Goto 3.