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Arrays: Practice Problems II - Reversing an Array In Place

Reading · 7 minQuiz · 5 questions2 code drills · run onlyGenerated by gpt-5.6-luna · Aug 27

Every index has a mirror at n - 1 - i

To reverse [4, 2, 7, 1, 5], each value must move to the index equally far from the other end. With n = 5, index 0 has mirror index 5 - 1 - 0 = 4, so 4 and 5 exchange positions. Index 1 has mirror index 5 - 1 - 1 = 3, so 2 and 1 exchange positions. Index 2 mirrors itself because 5 - 1 - 2 = 2, so 7 stays in the middle.

JAVAThe reversal pairs each index with n - 1 - i.
int[] a = {4, 2, 7, 1, 5};
int n = a.length;

// Required exchanges:
// index 0 with index 4
// index 1 with index 3
// index 2 stays where it is
The mirror pairs in [4, 2, 7, 1, 5]Five cells contain [4, 2, 7, 1, 5] at indices 0 through 4. A connection pairs index 0, value 4, with index 4, value 5. Another connection pairs index 1, value 2, with index 3, value 1. Index 2, value 7, is marked as its own mirror and remains unchanged.0123442715mirror pairmirror pairself-mirror: 7 staysi = 0: n - 1 - i = 4i = 4: n - 1 - i = 0Reverse swaps (0, 4) and (1, 3); index 2 is unchanged.
Reversal exchanges two mirror pairs and leaves the middle slot unchanged.

Direct assignment destroys a value that the next step still needs

The assignment a[i] = a[n - 1 - i] copies the mirror value into the left slot, but it does not preserve the value that was already there. For i = 0, the assignment copies 5 into index 0, changing the array from [4, 2, 7, 1, 5] to [5, 2, 7, 1, 5]. The original 4 is gone from the array. No later read of index 0 can recover it.

Continuing the same assignment for i = 1 copies 1 into index 1, changing the array to [5, 1, 7, 1, 5]. The original 2 is now gone as well. Later assignments cannot repair either loss, because they read values from the already changed array.

JAVAThis direct assignment does not reverse the array correctly because each left value is overwritten.
for (int i = 0; i < n; i++) {
    a[i] = a[n - 1 - i];
}
CHECKPOINT 1Not answered

What array results from running a[i] = a[n - 1 - i] for i = 0 through 4 on [4, 2, 7, 1, 5]?

Checkpoints are not graded. They are here so you catch yourself before the quiz does — stuck, ask the tutor on the right.

A temporary variable preserves both values during a swap

A swap must save the left value before either array slot is overwritten. For indices 0 and 4, temp first stores 4. Then a[0] receives 5, while temp still holds 4. Finally, a[4] receives temp. The array becomes [5, 2, 7, 1, 4], so both values have changed positions instead of one being lost.

JAVAThe three assignments swap the values at indices 0 and 4 without losing either value.
int i = 0;
int mirror = n - 1 - i;
int temp = a[i];
a[i] = a[mirror];
a[mirror] = temp;

The same order swaps indices 1 and 3. Before the swap, the array is [5, 2, 7, 1, 4]. Saving a[1] stores 2, copying a[3] into a[1] produces [5, 1, 7, 1, 4], and copying the saved value into a[3] produces [5, 1, 7, 2, 4].

The three assignments that swap indices 0 and 4The trace begins with [4, 2, 7, 1, 5], i equal to 0, and mirror index 4. The value 4 from index 0 is stored in temp. Copying the value 5 from index 4 into index 0 changes the array to [5, 2, 7, 1, 5], but temp still contains 4. Copying temp into index 4 completes the swap and produces [5, 2, 7, 1, 4].427154271552715527141 Starting state2 Save the value3 Left write4 Right writei = 0mirror = 4temp = 4a[0] = a[4]temp still holds 4a[4] = temp
The temporary variable keeps the overwritten value available.
CHECKPOINT 2Not answered

Complete the three assignment statements that swap indices 1 and 3 in [5, 2, 7, 1, 4]. Use temp as the temporary variable.

Checkpoints are not graded. They are here so you catch yourself before the quiz does — stuck, ask the tutor on the right.

Stopping at n / 2 prevents every swap from being undone

The array has only two distinct mirror pairs: indices 0 and 4, then indices 1 and 3. Index 2 is its own mirror and needs no change. Therefore the loop should run for i = 0 and i = 1, which is exactly i < n / 2 because integer division makes 5 / 2 equal to 2.

JAVAOnly the first half of the indices starts each mirror swap.
for (int i = 0; i < n / 2; i++) {
    int mirror = n - 1 - i;
    int temp = a[i];
    a[i] = a[mirror];
    a[mirror] = temp;
}

After i = 0, the array is [5, 2, 7, 1, 4]. After i = 1, it is [5, 1, 7, 2, 4]. If the loop continued through the second half, i = 3 would swap indices 3 and 1 again, and i = 4 would swap indices 4 and 0 again. Those repeated swaps would restore the original array instead of reversing it.

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