06:16
A Meera array is defined to be an array that
contains at least one odd number and begins and ends with the same number
of even numbers.
So {4, 8, 6, 3, 2, 9, 8,11, 8, 13, 12, 12, 6} is a Meera array because it begins with three even numbers and ends with three even numbers and it contains at least one odd number
The array {2, 4, 6, 8, 6} is not a Meera array because it does not contain an odd number.
The array {2, 8, 7, 10, -4, 6} is not a Meera array because it begins with two even numbers but ends with three even numbers.
Write a function named isMeera that returns 1 if its array argument is a Meera array. Otherwise, it returns 0.
If you are writing in Java or C#, the function signature is
int isMeera (int[ ] a)
If you are writing in C or C++, the function signature is
int isMeera (int a[ ], int len) where len is the number of elements in the array.
So {4, 8, 6, 3, 2, 9, 8,11, 8, 13, 12, 12, 6} is a Meera array because it begins with three even numbers and ends with three even numbers and it contains at least one odd number
The array {2, 4, 6, 8, 6} is not a Meera array because it does not contain an odd number.
The array {2, 8, 7, 10, -4, 6} is not a Meera array because it begins with two even numbers but ends with three even numbers.
Write a function named isMeera that returns 1 if its array argument is a Meera array. Otherwise, it returns 0.
If you are writing in Java or C#, the function signature is
int isMeera (int[ ] a)
If you are writing in C or C++, the function signature is
int isMeera (int a[ ], int len) where len is the number of elements in the array.
public static int
isMeera(int[] a)
{
bool isOdd = false, firstCountEnd = false, lastCountEnd = false;
int
firstCount = 0, lastCount = 0;
for (int i = 0; i < a.Length; i++)
{
if (a[i] %
2 != 0)
{
isOdd = true;
break;
}
}
for (int i = 0; i < a.Length; i++)
{
if (a[i] %
2 != 0)
firstCountEnd = true;
if
(!firstCountEnd && a[i] % 2 == 0)
firstCount++;
if
(a[(a.Length - 1) - i] % 2 != 0)
lastCountEnd = true;
if
(!lastCountEnd && a[(a.Length - 1) - i] % 2 == 0)
lastCount++;
}
if (isOdd
&& firstCount == lastCount)
{
return 1;
}
else
{
return 0;
}
}
# this is better code
ReplyDeletepublic static int meeraArray(int[] a){
int evenCount, oddNum, evenNum ;
evenCount = oddNum = evenNum = 0;
for (int i = 0; i < a.length; i++){
if(a[i] % 2 == 0 && oddNum == 0){
evenCount += 1;
evenNum = evenCount;
}
else if(a[i] % 2 != 0 && oddNum == 0){
oddNum += 1;
}
else if (a[i] % 2 == 0 && oddNum != 0){
if( i +1 < a.length && a[i+1] % 2 != 0){
evenCount = evenNum;
}
else{ evenCount -= 1;}
}
}
if (evenCount != 0 || oddNum ==0){ return 0;}
return 1;
}
This is an O(n) solution, let me know your thoughts. Thanks
ReplyDeletepublic static int isMeera(int [] a){
boolean foundOdd=false;
boolean countingEvens=true; //count evens and later odds
int evensCount=0;
for(int i=0;i0) return 1;
else return 0;
}
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