Answer:
Binary sort
Explanation:
Binary sort is one of the fastest search techniques used on arrays and other iterable data types. It algorithm sorts the array in ascending order then gets the item in the middle from which it divides the array to be searched in two parts. If the searched term is less than the mid item, then it is searched for in the first part, else it would be in the second.
Answer:
Going green has several other benefits for companies. These include tax credits and incentives, improved efficiency, healthier workplaces, and cost savings – for instance by printing less, turning lights off in unused rooms and refilling ink cartridges. Reusing items also reduces waste from plastic packaging.
Explanation:
Answer:
ICT is a great impact for are daily lives. It can improve the quality of a human life. The reason why it can improve that, it’s because it can be used as a learning and education media, the mass communication media in promoting and campaigning practical and important issues, such as the health and social area.
Explanation:
Review
(I was suprised too)
Answer:
Explanation:
The following program creates a function called region_Matches that takes in two strings as arguments as well as an int for starting point and an int for amount of characters to compare. Then it compares those characters in each of the words. If they match (ignoring case) then the function outputs True, else it ouputs False. The test cases compare the words "moving" and "loving", the first test case compares the words starting at point 0 which outputs false because m and l are different. Test case 2 ouputs True since it starts at point 1 which is o and o.
class Brainly {
public static void main(String[] args) {
String word1 = "moving";
String word2 = "loving";
boolean result = region_Matches(word1, word2, 0, 4);
boolean result2 = region_Matches(word1, word2, 1, 4);
System.out.println(result);
System.out.println(result2);
}
public static boolean region_Matches(String word1, String word2, int start, int numberOfChars) {
boolean same = true;
for (int x = 0; x < numberOfChars; x++) {
if (Character.toLowerCase(word1.charAt(start + x)) == Character.toLowerCase(word2.charAt(start + x))) {
continue;
} else {
same = false;
break;
}
}
return same;
}
}