Arrays (Continue) Sorting and searching

outlines Sorting – Bubble sort Linear search – min and max Binary search

Sort Bubble Sort

Sorting Sorting takes an unordered collection and makes it an ordered one

"Bubbling Up" the Largest Element Traverse a collection of elements – Move from the front to the end – Bubble the largest value to the end using pair- wise comparisons and swapping Swap 4277

"Bubbling Up" the Largest Element Traverse a collection of elements – Move from the front to the end – Bubble the largest value to the end using pair- wise comparisons and swapping Swap 3577

"Bubbling Up" the Largest Element Traverse a collection of elements – Move from the front to the end – Bubble the largest value to the end using pair- wise comparisons and swapping Swap 1277

"Bubbling Up" the Largest Element Traverse a collection of elements – Move from the front to the end – Bubble the largest value to the end using pair- wise comparisons and swapping No need to swap

"Bubbling Up" the Largest Element Traverse a collection of elements – Move from the front to the end – Bubble the largest value to the end using pair- wise comparisons and swapping Swap 5101

"Bubbling Up" the Largest Element Traverse a collection of elements – Move from the front to the end – Bubble the largest value to the end using pair- wise comparisons and swapping Largest value correctly placed

The Bubble Up Algorithm for(int i = 0; i<A.length; i++) { for(int j = 0; j < A.length-1; j++) { if(A[j]>A[j+1]) { temp= A[j]; A[j]=A[j+1]; A[j+1]=temp; }

Items of Interest Notice that only the largest value is correctly placed All other values are still out of order So we need to repeat this process Largest value correctly placed

Repeat Bubble Up How Many Times? If we have N elements… And if each time we bubble an element, we place it in its correct location… Then we repeat the bubble up process N – 1 times. This guarantees well correctly place all N elements.

Bubbling All the Elements N - 1

Reducing the Number of Comparisons

for(int i = 0; i<A.length; i++) { for(int j = 0; j < A.length-1; j++) { if(A[j]>A[j+1]) { temp= A[j]; A[j]=A[j+1]; A[j+1]=temp; }

Already Sorted Collections? What if the collection was already sorted? What if only a few elements were out of place and after a couple of bubble ups, the collection was sorted? We want to be able to detect this and stop early!

Using a Boolean Flag We can use a boolean variable to determine if any swapping occurred during the bubble up. If no swapping occurred, then we know that the collection is already sorted! This boolean flag needs to be reset after each bubble up.

Add flag to reduce the number of iterations Boolean swap = false; for(int i = 0; i<A.length && !swap; i++) { boolean swap = false; for(int j = 0; j < A.length-1; j++) { if(A[j]>A[j+1]) { temp= A[j]; A[j]=A[j+1]; A[j+1]=temp; swap = true; }

An Animated Example j index 0 A.length 8 swap true

An Animated Example to_do index 0 1 A.lenght 8 did_swap false

An Animated Example to_do index 0 1 N 8 Swap did_swap false

An Animated Example to_do index 0 1 N 8 Swap did_swap true

An Animated Example j index 7 2 A.length 8 did_swap true

An Animated Example to_do index 7 2 N 8 Swap did_swap true

An Animated Example j index 0 2 A.length 8 Swap did_swap true

An Animated Example j index 0 3 A.length 8 swap true

An Animated Example j index 0 3 A.lengt 8 Swap swap true

An Animated Example j index 0 3 A.lengt 8 Swap swap true

An Animated Example j index 0 4 A.lengt 8 swap true

An Animated Example j index 0 4 A.length 8 Swap did_swap true

An Animated Example j index 0 4 A.length 8 Swap did_swap true

An Animated Example j index 0 5 A.lengt 8 swap true

An Animated Example j index 0 5 A.lengt 8 Swap swap true

An Animated Example j index 0 5 A.length 8 Swap did_swap true

An Animated Example j index 0 6 A.length 8 swap true

An Animated Example j index 0 6 A.lengt 8 Swap swap true

An Animated Example j index 0 6 A.lengt 8 Swap swap true

An Animated Example j index 0 7 A.lengt 8 swap true

An Animated Example j index 0 7 A.length 8 Swap swap true

An Animated Example j index 0 7 A.length 8 Swap swap true

After First Pass of Outer Loop j index 0 8 A.length 8 Finished first Bubble Up swap true

The Second Bubble Up j index 2 1 A.length 8 swap false

The Second Bubble Up j index 2 1 A.length 8 swap false No Swap

The Second Bubble Up j index 1 2 A.length 8 swap false

The Second Bubble Up j index 1 2 A.length 8 did_swap false Swap

The Second Bubble Up j index 1 2 A.length 8 did_swap true Swap

The Second Bubble Up j index 1 3 A.length 8 swap true

The Second Bubble Up j index 1 3 A.length 8 did_swap true Swap

The Second Bubble Up j index 1 3 A.length 8 swap true Swap

The Second Bubble Up j index 1 4 A.length 8 did_swap true

The Second Bubble Up to_do index 1 4 N 8 swap true No Swap

The Second Bubble Up to_do index 1 5 N 8 swap true

The Second Bubble Up to_do index 1 5 N 8 swap true Swap

The Second Bubble Up j index 1 5 A.length 8 swap true Swap

The Second Bubble Up j index 1 6 A.length 8 swap true

The Second Bubble Up to_do index 1 6 N 8 swap true Swap

The Second Bubble Up to_do index 1 6 N 8 swap true Swap

After Second Pass of Outer Loop index 1 7 N 8 swap true Finished second Bubble Up

The Third Bubble Up j index 2 1 A.length 8 swap false

The Third Bubble Up j index 2 1 A.length 8 swap false Swap

The Third Bubble Up j index 2 1 A.length 8 swap true Swap

The Third Bubble Up j index 2 2 A.length 8 swap true

The Third Bubble Up j index 2 2 A.lengt 8 swap true Swap

The Third Bubble Up j index 2 2 A.lengt 8 swap true Swap

The Third Bubble Up to_do index 2 3 N 8 did_swap true

The Third Bubble Up j index 2 3 A.length 8 swap true No Swap

The Third Bubble Up j index 2 4 A.length 8 did_swap true

The Third Bubble Up j index 2 4 A.length 8 swap true Swap

The Third Bubble Up j index 2 4 A.length 8 swap true Swap

The Third Bubble Up j index 2 5 A.length 8 did_swap true

The Third Bubble Up j index 2 5 A.length 8 swap true Swap

The Third Bubble Up j index 2 5 A.length 8 did_swap true Swap

After Third Pass of Outer Loop j index 2 6 A.length 8 did_swap true Finished third Bubble Up

The Fourth Bubble Up j index 3 1 A.length 8 swap false

The Fourth Bubble Up j index 3 1 A.length 8 swap false Swap

The Fourth Bubble Up to_do index 3 1 N 8 did_swap true Swap

The Fourth Bubble Up j index 3 2 A.length 8 swap true

The Fourth Bubble Up j index 3 2 A.length 8 did_swap true No Swap

The Fourth Bubble Up j index 3 3 A.length 8 swap true

The Fourth Bubble Up j index 3 3 A.length 8 did_swap true No Swap

The Fourth Bubble Up j index 3 4 A.length 8 swap true

The Fourth Bubble Up j index 3 4 A.length 8 did_swap true No Swap

After Fourth Pass of Outer Loop j index 3 5 A.length 8 swap true Finished fourth Bubble Up

The Fifth Bubble Up j index 4 1 A.length 8 swap false

The Fifth Bubble Up j index 4 1 A.lenght 8 swap false No Swap

The Fifth Bubble Up j index 4 2 A.length 8 swap false

The Fifth Bubble Up j index 4 2 A.length 8 swap false No Swap

The Fifth Bubble Up j index 4 3 A.length 8 swap false

Finished Early j index 3 4 A.lengt 8 swap false We didnt do any swapping, so all of the other elements must be correctly placed. We can skip the last two passes of the outer loop.

Bubble sort Summary Bubble Up algorithm will move largest value to its correct location (to the right) Repeat Bubble Up until all elements are correctly placed: – Maximum of N-1 times – Can finish early if no swapping occurs We reduce the number of elements we compare each time one is correctly placed

Linear search Use linear to seach and value is contained in one array.

Searching for x in an array a for(int i = 0; i < a.length; ++ i) { if (a[i] == x) System.out.println(true); }

Finding the minimal value in array int minValue = a[0]; int minIndex = 0; for(int i = start + 1; i < end; i ++) { if (a[i] < minValue) { minValue = a[i]; minIndex = i; }

Finding the minimal value in array int maxValue = a[0]; int maxIndex = 0; for(int i = start + 1; i < end; i ++) { if (a[i] > maxValue) { maxValue = a[i]; maxIndex = i; }

Binary search for a value in array Searching if x is in array a. We assume a is sorted in ascending order. int start = 0; int end = a.length - 1; boulean found = flase while (start <= end) { int m = (start + end)/2; if (x == a[m]) {found =false;} else if (x < a[m]) { end = m - 1;} else { start = m + 1; }

summary Buble sort Learnear search Binary sort