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Create Subtract Operations Explanation.txt

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2022/Contests/Combined Divisions/CodeTON/Explanations
- Subtract Operations Explanation.txt

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	`1`	`+Suppose we subtract the whole array by X in the first step`
	`2`	`+`
	`3`	`+Each A2[i] = A[i] - x`
	`4`	`+`
	`5`	`+Suppose we subtract the whole array by Y = A2[j] in the second step`
	`6`	`+`
	`7`	`+Each A3[i] = A2[i] - Y = (A[i] - X) - (A[j] - X)`
	`8`	`+`
	`9`	`+Where A[i], A2[i] and A3[i] represent the state of A[i] at time 1, 2, 3 respectively`
	`10`	`+`
	`11`	`+X gets cancelled out twice.`
	`12`	`+`
	`13`	`+Every operation cancels out everything that was subtracted in the previous step.`
	`14`	`+`
	`15`	`+What we will be left with in the end is the difference of 2 elements A[i] - A[j]`
	`16`	`+`
	`17`	`+We have to check if there is a pair who's difference is K in the original array.`
	`18`	`+`
	`19`	`+------`
	`20`	`+`
	`21`	`+void solve()`
	`22`	`+{`
	`23`	`+ int no_of_elements, k;`
	`24`	`+ cin >> no_of_elements >> k;`
	`25`	`+`
	`26`	`+ vector <long long> A(no_of_elements + 1);`
	`27`	`+ for(int i = 1; i <= no_of_elements; i++)`
	`28`	`+ {`
	`29`	`+ cin >> A[i];`
	`30`	`+ }`
	`31`	`+`
	`32`	`+ map <int, int> present;`
	`33`	`+ for(int i = 1; i <= no_of_elements; i++)`
	`34`	`+ {`
	`35`	`+ present[A[i]] = true;`
	`36`	`+ }`
	`37`	`+`
	`38`	`+ int possible = false;`
	`39`	`+ for(int i = 1; i <= no_of_elements; i++)`
	`40`	`+ {`
	`41`	`+ if(present[A[i] - k] \|\| present[A[i] + k])`
	`42`	`+ {`
	`43`	`+ possible = true;`
	`44`	`+ break;`
	`45`	`+ }`
	`46`	`+ }`
	`47`	`+`
	`48`	`+ cout << (possible ? "Yes" : "No") << "\n";`
	`49`	`+}`

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