Created
January 24, 2017 22:17
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Multiply Complexities
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// Computers m binary searches | |
public static boolean containsElements(int[] seaches, int[] array) | |
{ | |
boolean hasElements = true; | |
int m = searches.length; | |
int n = array.length; | |
// O(m) loop | |
for(int i = 0; i < m; i++) { | |
// Each Binary Search has O(lg n) complexity | |
if(binarySearch(array, searches[i] == false) | |
hasElements = false; | |
} | |
// Now for each O(m) we are performing a O(lg n) operation | |
// Therefore we O(m * lg n) | |
return hasElements; | |
} | |
// Brute force method to check if an array has unique elements | |
public static boolean containsUniqueElements(int[] array) | |
{ | |
int n = array.length; | |
boolean hasUniqueElements = true; | |
// O(n) Outer for loop. | |
for(int i = 0; i < n; i++) { | |
// O(n) Inner for loop. | |
for(int j = 0; j < n; j++) { | |
if(i != j && a[i] == a[j]) | |
hasUniqueElements = fasle; | |
} | |
} | |
/* For each of the O(n) outer loop operations we are performing | |
another O(n) operations which makes this O(n * n) = O (n ^ 2) | |
*/ | |
return hasUniqueElements; | |
} |
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