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# Orhan Obutorhanobut

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Last active Dec 23, 2015
Remove duplicates from an integer array and return the array with original order with only primitives. Complexity = O(n^2)
View gist:6637712
 public static int[] removeDuplicates(int[] list) { int count = 0; for (int i = 0; i < list.length; i++) { boolean isExist = false; int value = list[i]; list[i] = 0; for (int j = 0; j < count; j++) { if (list[j] == value) { isExist = true; break;
Last active Dec 23, 2015
Remove duplicated values from an integer array with using sorting. Complexity = O (n.log(n))
View gist:6637735
 public static int[] sortAndRemove(int[] list) { list = mergeSort(list); int[] temp = new int[list.length]; temp[0] = list[0]; int j = 1; for (int i = 1; i < list.length; i++) { if (list[i] != list[i - 1]) { temp[j] = list[i]; j++; }
Last active Dec 23, 2015
Fibonacci with recursive. Complexity = O (n!)
View gist:6637742
 public static int fib(int n) { if (n > 1) { return fib(n - 1) + fib(n - 2); } return n; }
Last active Dec 23, 2015
Factorial with recursive. Complexity = O (n!)
View gist:6637748
 public static int fact(int n) { if (n > 1) { return n * fact(n - 1); } return n; }
Last active Dec 23, 2015
Reverse a string in-space without an extra buffer. Complexity = O (n)
View gist:6637761
 public static String reverse(String s) { int lenght = s.length(); int last = s.length() - 1; char[] temp = s.toCharArray(); for (int i = 0; i < lenght / 2; i++) { char c = temp[i]; temp[i] = temp[last - i]; temp[last - i] = c; } return new String(temp);
Last active Dec 23, 2015
Remove duplicate char from a string value. Complexity = O (n^2)
View gist:6637780
 public static String removeDuplicateChars(String s) { if (s == null) return s; int len = s.length(); if (len < 2) { return s; } char[] chars = s.toCharArray(); int index = 0; for (int i = 0; i < len; i++) {
Last active Dec 23, 2015
Remove duplicate chars from a string for english word. Complexity = O (n)
View gist:6637791
 public static String removeDuplicateChars2(String s) { if (s == null) { return s; } int len = s.length(); if (len < 2){ return s; }
Last active Dec 23, 2015
Algorithm to check if the given 2 strings are anagram. Complexity = O (n)
View gist:6637803
 public static boolean checkIfAnagram(String s1, String s2) { int len1 = s1.length(); for (int i = 0; i < len1; i++) { if (s1.charAt(i) != s2.charAt(len1 - i - 1)) { return false; } } return true; }
Last active Dec 23, 2015
Get the minimum value of a given integer array. Complexity = O (n)
View gist:6637818
 public static int getMin(int[] a) { if (a == null) { return 0; } if (a.length == 1) { return a[0]; } int min = a[0]; for (int i = 0; i < a.length; i++) { if (a[i] < min) {
Last active Dec 23, 2015
Get the minimum value of a given sorted integer array. Complexity log(n)
View gist:6637825
 public static int getMinValue(int[] a) { if (a == null) return -1; int lo = 0; int hi = a.length - 1; while (lo <= hi) { int mid = lo + (hi - lo) / 2; if (a[mid] > a[hi]) { lo = mid + 1; } else if (a[mid] < a[hi]) {
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