ARRAY_IMPL1

Predict the output of the following program :


int* performOps(int *A, int len, int *blen) {
    int i;
    *blen = len * 2;
    int *B = (int *)malloc((*blen) * sizeof(int));
    for (i = 0; i < len; i++) {
        B[i] = A[i];
        B[i + len] = A[(len - i) % len];
    }
    return B;
}

Lets say performOps was called with len = 4, and A : [5, 10, 2, 1].

What would be the output of the following call :

( NOTE : The output shoudl be written in the below provided text field , there should not be any ‘,’ between the numbers . For example: 1 2 3 4 )


int blen; 
int *B = performOps(A, len, &blen);
int i;
for (i = 0; i < blen; i++) {
    printf("%d ", B[i]);
}


vector<int> performOps(vector<int> A) {
    vector<int> B(2 * A.size(), 0);
    for (int i = 0; i < A.size(); i++) {
        B[i] = A[i];
        B[i + A.size()] = A[(A.size() - i) % A.size()];
    }
    return B;
}

Lets say performOps was called with A : [5, 10, 2, 1]. What would be the output of the following call :

( NOTE : The output shoudl be written in the below provided text field , there should not be any ‘,’ between the numbers . For example: 1 2 3 4 )


vector<int> B = performOps(A);
for (int i = 0; i < B.size(); i++) {
    cout<<B[i]<<" ";
}


ArrayList<Integer> performOps(ArrayList<Integer> A) {
        ArrayList<Integer> B = new ArrayList<Integer>();
        for (int i = 0; i < 2 * A.size(); i++) B.add(0);
        for (int i = 0; i < A.size(); i++) {
                B.set(i, A.get(i));
                B.set(i + A.size(), A.get((A.size() - i) % A.size()));
        }
        return B;
}

Lets say performOps was called with A : [5, 10, 2, 1]. What would be the output of the following call :

( NOTE : The output shoudl be written in the below provided text field , there should not be any ‘,’ between the numbers . For example: 1 2 3 4 )


ArrayList<Integer> B = performOps(A);
for (int i = 0; i < B.size(); i++) {
        System.out.print(B.get(i) + " ");
}

def performOps(A):
    blen = 2 * len(A)
    B = [0]*blen
    for i in xrange(len(A)):
        B[i] = A[i]
        B[i + len(A)] = A[(len(A) - i) % len(A)]
    return B

Lets say performOps was called with A : [5, 10, 2, 1]. What would be the output of the following call :

( NOTE : The output shoudl be written in the below provided text field , there should not be any ‘,’ between the numbers . For example: 1 2 3 4 )

B = performOps(A)
for i in xrange(len(B)):
    print B[i],

function performOps(A){
    B = new Array(2 * A.length)

    for (var i = 0; i < A.length; i++) {
        B[i] = A[i];
        B[i + A.length] = A[(A.length - i) % A.length];
    }
    return B;
}

Lets say performOps was called with A : [5, 10, 2, 1]. What would be the output of the following call :

( NOTE : The output shoudl be written in the below provided text field , there should not be any ‘,’ between the numbers . For example: 1 2 3 4 )

B = performOps(A)
for (var i = 0; i < B.length; i++) {
    process.stdout.write(B[i]+" ");
}

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ARRAY_IMPL1

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