added array-smallest-missing-number

array-minimum-distance/array-minimum-distance.go

@@ -29,11 +29,9 @@ func minDist (arr []int, n, x, y int) int {
 
 	for i := 0; i < n; i++ {
 		for j := i + 1; j < n; j++ {
-
-			if ((x == arr[i] && y == arr[j] || y == arr[i] && x == arr[j]) && min_dist > Abs(i - j)) {
+			if((x == arr[i] && y == arr[j] || y == arr[i] && x == arr[j]) && min_dist > Abs(i - j)) {
 				min_dist = Abs(i - j)
 			}
-			
 		}
 	}
 

array-smallest-missing-number/README.md

@@ -0,0 +1,64 @@
+<h1 align="center">Array Smallest Missing Number Source</h1>
+
+[What It Is](#what-it-is)
+
+## What It Is
+
+Given a sorted array of `n` distinct integers where each integer is in the range from `0 to m-1 and m > n`. Find the smallest number that is missing from the array.
+
+Examples
+--------------------------
+
+> * Input: {0, 1, 2, 6, 9}, n = 5, m = 10
+> * Output: `3`
+
+> * Input: {4, 5, 10, 11}, n = 4, m = 12
+> * Output: `0`
+
+> * Input: {0, 1, 2, 3}, n = 4, m = 5
+> * Output: `4`
+
+> * Input: {0, 1, 2, 3, 4, 5, 6, 7, 10}, n = 9, m = 11
+> * Output: `8`
+
+**METHOD 1 (Use Binary Search)**
+
+`For i = 0 to m-1`, do binary search for i in the array. If i is not present in the array then `return i`.
+
+Algorithm Complexity
+--------------------------
+
+| Complexity	    | Notation     |
+| ----------------- |:------------:|
+| `Time Complexity`	| `O(m log n)` |
+
+**METHOD 2 (Linear Search)**
+
+If `arr[0] is not 0`, `return 0`. Otherwise traverse the input array starting from index 0, and for each pair of elements a[i] and a[i+1], find the difference between them. if the difference is greater than 1 then `a[i] + 1` is the missing number.
+
+Algorithm Complexity
+--------------------------
+
+| Complexity	    | Notation     |
+| ----------------- |:------------:|
+| `Time Complexity`	| `O(n)`       |
+
+
+**METHOD 3 (Use Modified Binary Search)**
+
+In the standard Binary Search process, the element to be searched is compared with the middle element and on the basis of comparison result, we decide whether to search is over or to go to left half or right half.
+In this method, we modify the standard Binary Search algorithm to compare the middle element with its index and make decision on the basis of this comparison.
+
+1) If the first element is not same as its index then return first index
+2) Else get the middle index say mid
+    a) If arr[mid] greater than mid then the required element lies in left half.
+    b) Else the required element lies in right half.
+
+`Note: This method doesn’t work if there are duplicate elements in the array.`
+
+Algorithm Complexity
+--------------------------
+
+| Complexity	    | Notation     |
+| ----------------- |:------------:|
+| `Time Complexity`	| `O(log n)`   |

array-smallest-missing-number/array-smallest-missing-number.go

@@ -0,0 +1,36 @@
+// ====================================================
+// Data-Structures-with-Go Copyright(C) 2017 Furkan Türkal
+// This program comes with ABSOLUTELY NO WARRANTY; This is free software,
+// and you are welcome to redistribute it under certain conditions; See
+// file LICENSE, which is part of this source code package, for details.
+// ====================================================
+
+package main
+
+import "fmt"
+
+func findFirstMissing (arr []int, start, end int) int {
+	
+	if(start > end) {
+		return end + 1
+	}
+
+	if(start != arr[start]) {
+		return start
+	}
+
+	mid := (start + end) / 2
+
+	if(arr[mid] == mid){
+		return findFirstMissing(arr, mid + 1, end)
+	}
+
+	return findFirstMissing(arr, start, end)
+}
+
+
+func main() {
+	arr := []int{0, 1, 2, 3, 4, 5, 6, 7, 10}
+	var n int = len(arr)
+	fmt.Printf("Smallest missing element is %d", findFirstMissing(arr, 0, n - 1))
+}