Writing a program in C++ that combines the three summations and linear search algorithm together in one code and there will be an improvement in the three summation algorithm performance? three...


Writing a program in C++ that combines the three summations and linear search algorithm<br>together in one code and there will be an improvement in the three summation algorithm<br>performance?<br>three summations<br>3-sum: brute-force algorithm<br>Linear Search Algorithm<br>publie elass Threesun<br>• The linear search algorithm locates an item in a list by examining elements in<br>the sequence one at a time, starting at the beginning.<br>• First compare x with a,. If they are equal, return the position 1.<br>• If not, try a,, If x = a, return the position 2.<br>• Keep going, and if no match is found when the entire list is scanned,<br>publie statie int count (int(] a)<br>int N-a.length:<br>int count- o:<br>for (int i- 0; i<N; 1++)<br>for (int j = 1+1: 3< N; ++)<br>for (int k- j+1; k< N; k++)<br>if (a(4) + al1 + a[k] - 0)<br>return 0.<br>check each triple<br>procedure linear search(x:integer,<br>a, a, ..a: distinct integers)<br>for simplicity, ignore<br>count++:<br>integer overflow<br>return count;<br>i:-1<br>while (isn and x + a)<br>public static void nain (String() args)<br>i:=i+1<br>ifisn then location := i<br>else location := 0<br>return location{location is the subscript of the term that<br>equals x, or is 0 if x is not found}<br>int() a- In. readints (arga (0]):<br>Stdout.printin (count (a) ):<br>

Extracted text: Writing a program in C++ that combines the three summations and linear search algorithm together in one code and there will be an improvement in the three summation algorithm performance? three summations 3-sum: brute-force algorithm Linear Search Algorithm publie elass Threesun • The linear search algorithm locates an item in a list by examining elements in the sequence one at a time, starting at the beginning. • First compare x with a,. If they are equal, return the position 1. • If not, try a,, If x = a, return the position 2. • Keep going, and if no match is found when the entire list is scanned, publie statie int count (int(] a) int N-a.length: int count- o: for (int i- 0; i< n;="" ++)="" for="" (int="" k-="" j+1;="">< n;="" k++)="" if="" (a(4)="" +="" al1="" +="" a[k]="" -="" 0)="" return="" 0.="" check="" each="" triple="" procedure="" linear="" search(x:integer,="" a,="" a,="" ..a:="" distinct="" integers)="" for="" simplicity,="" ignore="" count++:="" integer="" overflow="" return="" count;="" i:-1="" while="" (isn="" and="" x="" +="" a)="" public="" static="" void="" nain="" (string()="" args)="" i:="i+1" ifisn="" then="" location="" :="i" else="" location="" :="0" return="" location{location="" is="" the="" subscript="" of="" the="" term="" that="" equals="" x,="" or="" is="" 0="" if="" x="" is="" not="" found}="" int()="" a-="" in.="" readints="" (arga="" (0]):="" stdout.printin="" (count="" (a)="">

Jun 01, 2022
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