Write the state of the elements of each of the following arrays after each pass of the outermost loop of the selection sort algo

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Write the state of the elements of each of the following arrays after each pass of the outermost loop of the selection sort algo

rithm has occurred (after each element is selected and moved into place). int[] numbers3 = {8, 5, –9, 14, 0, –1, –7, 3}; int[] numbers4 = {15, 56, 24, 5, 39, –4, 27, 10};

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1 answer:

Natasha_Volkova [1K] · Answer 1 · 2026-04-08T06:51:01+03:00

Answer:

Below is the explanation provided.

Elaboration:

In the first array => 8, 5, -9, 14, 0, -1, -7, 3.

8, 5, -9, 14, 0, -1, -7, 3 For the first iteration, find the smallest from the first to the eighth and reposition the third element in the first place.
-9, 8, 5, 14, 0, -1, -7, 3 In the second round, identify the smallest from the second to eighth and adjust the penultimate element in the second slot.
-9, -7, 8, 5, 14, 0, -1, 3 For the third pass, find the least from third to eighth and move the second last item from that segment.
-9, -7, -1, 8, 5, 14, 0, 3 For the fourth pass, find the minimum from the fourth to eighth and shift the last second item from that section.
-9, -7, -1, 0, 8, 5, 14, 3 For the fifth pass, seek for the smallest from the fifth to eighth and place the last element from that portion.
-9, -7, -1, 0, 3, 8, 5, 14 For the sixth pass, identify the smallest from the sixth to eighth and shift the last second element from that group.
-9, -7, -1, 0, 3, 5, 8, 14 are in order now.

In the second array => 15, 56, 24, 5, 39, -4, 27, 10.

15, 56, 24, 5, 39, -4, 27, 10 In the first iteration, identify the smallest item from the first to the eighth and reposition the last third item in that section.
-4, 15, 56, 24, 5, 39, 27, 10 In the second round, find the minimum item from second to eighth and interchange the last fourth item in that segment.

-4, 5, 15, 56, 24, 39, 27, 10 For the third pass, select the smallest from the third to eighth and shift the last item from that set.
-4, 5, 10, 15, 56, 24, 39, 27 For the fourth stage, determine the smallest from the fourth to eighth and do not shift its position as it is already sorted.
-4, 5, 10, 15, 56, 24, 39, 27 For the fifth iteration, check for the smallest from the fifth to the eighth and intersperse the last third item from that section.

-4, 5, 10, 15, 24, 56, 39, 27 For the sixth phase, find the smallest from the sixth to the eighth and exchange the last item in the sixth.

-4, 5, 10, 15, 24, 27, 56, 39 are in order now.