Binary Search
Prateek Narang
Binary Search
What you will learn?
-
Binary Search Concept
-
Code
-
Complexity Analysis
Searching is an important operation in every-day life and even in many software applications, as we frequently search for data!
Searching is a method to find some relevant information in a data set
Searching
0 1 2 3 4 5 6 7 8
Lets try to search for Number 12 in an array.
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0 1 2 3 4 5 6 7 8
Lets try to search for Number 12 in an array.
1 3 4 5 9 12 22 38 45
12
0 1 2 3 4 5 6 7 8
Lets try to search for Number 12 in an array.
1 3 4 5 9 12 22 38 45
12
0 1 2 3 4 5 6 7 8
Lets try to search for Number 12 in an array.
1 3 4 5 9 12 22 38 45
12
0 1 2 3 4 5 6 7 8
Lets try to search for Number 12 in an array.
1 3 4 5 9 12 22 38 45
12
0 1 2 3 4 5 6 7 8
Lets try to search for Number 12 in an array.
1 3 4 5 9 12 22 38 45
12
0 1 2 3 4 5 6 7 8
Lets try to search for Number 12 in an array.
1 3 4 5 9 12 22 38 45
12
0 1 2 3 4 5 6 7 8
Lets try to search for Number 12 in an array.
1 3 4 5 9 12 22 38 45
12
and there is a Match at index 5!
This algorithm of linearly searching through the array is known as
Linear Search!
-
Can we do better?
-
Can we exploit the information that the array is sorted?
Binary Search is an algorithm that can be used to search an element in a sorted dataset.
By sorted, we mean that the elements will either be in a natural increasing or decreasing order.
🚀 Binary Search
10, 20, 28 , 56, 100
(Sorted Array)
0 1 2 3 4 5 6 7 8
Lets try to search for Number 12 in an array.
s
e
Initialise
s = 0, e = 8
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0 1 2 3 4 5 6 7 8
Lets try to search for Number 12 in an array.
s
e
Find the midpoint
s = 0, e = 8
mid = (s+e)/2
= (0+8)/2
= 4
mid
1 3 4 5 9 12 22 38 45
0 1 2 3 4 5 6 7 8
Lets try to search for Number 12 in an array.
s
e
mid
Make a comparison between
arr[mid] & key
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12
9 <
0 1 2 3 4 5 6 7 8
s
e
mid
9 <
Clearly
So, the we discard the part of the array from s to mid.
12
1 3 4 5 9 12 22 38 45
0 1 2 3 4 5 6 7 8
s
e
1 3 4 5 9 12 22 38 45
s = mid + 1
Hence, the new start becomes 5.
mid
0 1 2 3 4 5 6 7 8
s
e
mid
mid = (s + e)/2
= (5 + 8)/2
= 6
1 3 4 5 9 12 22 38 45
0 1 2 3 4 5 6 7 8
s
e
mid
Now again, we compare a[mid] and key
22 >
12
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0 1 2 3 4 5 6 7 8
s
e
mid
a[mid] >
key
1 3 4 5 9 12 22 38 45
e = mid - 1
0 1 2 3 4 5 6 7 8
s
e
mid = (5 + 5)/2
= 5
mid
1 3 4 5 9 12 22 38 45
0 1 2 3 4 5 6 7 8
s
e
mid
a[mid] ==
12
1 3 4 5 9 12 22 38 45
0 1 2 3 4 5 6 7 8
s
e
mid
a[mid] ==
return mid;
12
1 3 4 5 9 12 22 38 45
0 1 2 3 4 5 6 7 8
s
e
mid
and we got the answer that the final index of 12 is actually 5.
1 3 4 5 9 12 22 38 45
What If, the element is
NOT PRESENT 🧐
0 1 2 3 4 5 6 7 8
Lets try to search for Number 6 in an array.
s
e
1 3 4 5 9 12 22 38 45
0 1 2 3 4 5 6 7 8
Lets try to search for Number 6 in an array.
s
e
mid
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mid = (0 + 8)/2 = 4
0 1 2 3 4 5 6 7 8
Lets try to search for Number 6 in an array.
s
e
mid
9 >
Search in the left of mid
6
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0 1 2 3 4 5 6 7 8
Lets try to search for Number 6 in an array.
s
e
mid
e = mid - 1
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0 1 2 3 4 5 6 7 8
Lets try to search for Number 6 in an array.
e
mid = (0 + 3)/2 = 1
mid
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s
0 1 2 3 4 5 6 7 8
Lets try to search for Number 6 in an array.
s
e
mid
Search in the right of mid
6
3 <
1 3 4 5 9 12 22 38 45
0 1 2 3 4 5 6 7 8
Lets try to search for Number 6 in an array.
s
e
mid
Search in the right of mid
6
3 <
1 3 4 5 9 12 22 38 45
0 1 2 3 4 5 6 7 8
Lets try to search for Number 6 in an array.
e
mid
s = mid + 1
6
3 <
s
1 3 4 5 9 12 22 38 45
0 1 2 3 4 5 6 7 8
Lets try to search for Number 6 in an array.
e
s = mid + 1
6
3 <
s
1 3 4 5 9 12 22 38 45
mid
0 1 2 3 4 5 6 7 8
Lets try to search for Number 6 in an array.
e
mid = (2 + 3)/2 = 2
s
mid
1 3 4 5 9 12 22 38 45
0 1 2 3 4 5 6 7 8
Lets try to search for Number 6 in an array.
e
Compare
s
mid
4 <
6
1 3 4 5 9 12 22 38 45
0 1 2 3 4 5 6 7 8
Lets try to search for Number 6 in an array.
e
Go Right, s = mid + 1
s
mid
4 <
6
1 3 4 5 9 12 22 38 45
0 1 2 3 4 5 6 7 8
Lets try to search for Number 6 in an array.
e
Go Right, s = mid + 1
s
mid
4 <
6
1 3 4 5 9 12 22 38 45
0 1 2 3 4 5 6 7 8
Lets try to search for Number 6 in an array.
e
s
mid
mid = (3 + 3)/2 = 3
1 3 4 5 9 12 22 38 45
0 1 2 3 4 5 6 7 8
Lets try to search for Number 6 in an array.
e
s
mid
Go right, s = mid + 1
5 <
6
1 3 4 5 9 12 22 38 45
0 1 2 3 4 5 6 7 8
Lets try to search for Number 6 in an array.
e
s
mid
Go right, s = mid + 1
5 <
6
1 3 4 5 9 12 22 38 45
0 1 2 3 4 5 6 7 8
Lets try to search for Number 6 in an array.
e
s
start becomes greater than end! 🧐
s > e
1 3 4 5 9 12 22 38 45
mid
0 1 2 3 4 5 6 7 8
Lets try to search for Number 6 in an array.
e
s
At this point, s > e, and it means we have covered the entire array but the element was not present,
and we STOP our search at this point.
1 3 4 5 9 12 22 38 45
Binary Search is a Divide & Conquer Algorithm.
Binary Search
It divides the array into two halves and tries to find the element K in one or the other half, not both. It keeps on doing this until we either find the element or the array is exhausted.
Code 👨💻
private static int binarySearch(int[] array, int K) {
}private static int binarySearch(int[] array, int K) {
int n = array.length;
int s = 0;
int e = n - 1;
}private static int binarySearch(int[] array, int K) {
int n = array.length;
int s = 0;
int e = n - 1;
while (s <= e) {
// Compute Mid
int mid = (s + e)/2;
}
}private static int binarySearch(int[] array, int K) {
int n = array.length;
int s = 0;
int e = n - 1;
while (s <= e) {
// Compute Mid
int mid = (s + e)/2;
// Go Right
if (array[mid] < K) {
s = mid + 1;
}
}
}private static int binarySearch(int[] array, int K) {
int n = array.length;
int s = 0;
int e = n - 1;
while (s <= e) {
// Compute Mid
int mid = (s + e)/2;
// Go Right
if (array[mid] < K) {
s = mid + 1;
}
// Go Left
else if (array[mid] > K) {
e = mid - 1;
}
}
}private static int binarySearch(int[] array, int K) {
int n = array.length;
int s = 0;
int e = n - 1;
while (s <= e) {
// Compute Mid
int mid = (s + e)/2;
// Go Right
if (array[mid] < K) {
s = mid + 1;
}
// Go Left
else if (array[mid] > K) {
e = mid - 1;
}
// Equal!
else {
return mid;
}
}
}private static int binarySearch(int[] array, int K) {
int n = array.length;
int s = 0;
int e = n - 1;
while (s <= e) {
// Compute Mid
int mid = (s + e)/2;
// Go Right
if (array[mid] < K) {
s = mid + 1;
}
// Go Left
else if (array[mid] > K) {
e = mid - 1;
}
// Equal!
else {
return mid;
}
}
return -1;
}Let's discuss space and time
Complexity Analysis
Time? 👨💻
Let's dicuss the worst complexity
1 3 4 5 9 12 22 38 45
1 3 4 5 9 12 22 38 45
1 3 4 5 9 12 22 38 45
Iteration , Array Size
1 N/2
2 N/4
3 N/8
.
.
Stop
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1 3 4 5 9 12 22 38 45
1 3 4 5 9 12 22 38 45
1 3 4 5 9 12 22 38 45
1 3 4 5 9 12 22 38 45
1 3 4 5 9 12 22 38 45
1 3 4 5 9 12 22 38 45
1 3 4 5 9 12 22 38 45
1 3 4 5 9 12 22 38 45
private static int binarySearch(int[] array, int K) {
int n = array.length;
int s = 0;
int e = n - 1;
while (s <= e) {
// Compute Mid
int mid = (s + e)/2;
// Go Right
if (array[mid] < K) {
s = mid + 1;
}
// Go Left
else if (array[mid] > K) {
e = mid - 1;
}
// Equal!
else {
return mid;
}
}
return -1;
}Let's try to generalise this behaviour
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1 3 4 5 9 12 22 38 45
1 3 4 5 9 12 22 38 45
1 3 4 5 9 12 22 38 45
1 3 4 5 9 12 22 38 45
1 3 4 5 9 12 22 38 45
1 3 4 5 9 12 22 38 45
1 3 4 5 9 12 22 38 45
1 3 4 5 9 12 22 38 45
1 3 4 5 9 12 22 38 45
1 3 4 5 9 12 22 38 45
1 3 4 5 9 12 22 38 45
private static int binarySearch(int[] array, int K) {
int n = array.length;
int s = 0;
int e = n - 1;
while (s <= e) {
// Compute Mid
int mid = (s + e)/2;
// Go Right
if (array[mid] < K) {
s = mid + 1;
}
// Go Left
else if (array[mid] > K) {
e = mid - 1;
}
// Equal!
else {
return mid;
}
}
return -1;
}Space? 👨💻
private static int binarySearch(int[] array, int K) {
int n = array.length;
int s = 0;
int e = n - 1;
while (s <= e) {
// Compute Mid
int mid = (s + e)/2;
// Go Right
if (array[mid] < K) {
s = mid + 1;
}
// Go Left
else if (array[mid] > K) {
e = mid - 1;
}
// Equal!
else {
return mid;
}
}
return -1;
}