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Two pointers method, слайд №1 Two pointers method, слайд №2 Two pointers method, слайд №3 Two pointers method, слайд №4 Two pointers method, слайд №5 Two pointers method, слайд №6 Two pointers method, слайд №7 Two pointers method, слайд №8 Two pointers method, слайд №9 Two pointers method, слайд №10 Two pointers method, слайд №11 Two pointers method, слайд №12 Two pointers method, слайд №13 Two pointers method, слайд №14 Two pointers method, слайд №15 Two pointers method, слайд №16 Two pointers method, слайд №17 Two pointers method, слайд №18 Two pointers method, слайд №19 Two pointers method, слайд №20 Two pointers method, слайд №21 Two pointers method, слайд №22 Two pointers method, слайд №23 Two pointers method, слайд №24 Two pointers method, слайд №25 Two pointers method, слайд №26 Two pointers method, слайд №27 Two pointers method, слайд №28 Two pointers method, слайд №29 Two pointers method, слайд №30 Two pointers method, слайд №31 Two pointers method, слайд №32 Two pointers method, слайд №33 Two pointers method, слайд №34 Two pointers method, слайд №35 Two pointers method, слайд №36 Two pointers method, слайд №37 Two pointers method, слайд №38
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TWO POINTERS METHOD Lyzhin Ivan, 2016
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TWO POINTERS METHOD Lyzhin Ivan, 2016

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Problem There is array A with N positive integers. Segment of array – a sequence of one or more consecutive elements in array. D-good segment –...
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Problem There is array A with N positive integers. Segment of array – a sequence of one or more consecutive elements in array. D-good segment – segment, in which sum of elements not greater than D. Count the pairs (L, R) such that segment [L, R] of array A is D-good.

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1. Very primitive solution Sum elements for each pair (L, R).
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1. Very primitive solution Sum elements for each pair (L, R).

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2. Primitive solution Notice that sum(L, R) = sum(L, R-1)+A[R] If sum(L, R1)>D then sum(L, R2)>D for each R2>R1
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2. Primitive solution Notice that sum(L, R) = sum(L, R-1)+A[R] If sum(L, R1)>D then sum(L, R2)>D for each R2>R1

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3. Good solution Notice that it’s enough to find maxR(L) = max(R) such sum(L, R)D for each R’>R. We can precompute prefix sums and than find maxR by...
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3. Good solution Notice that it’s enough to find maxR(L) = max(R) such sum(L, R)D for each R’>R. We can precompute prefix sums and than find maxR by binary search.

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3. Good solution
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3. Good solution

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4. Best solution Notice that maxR(L)>=maxR(L-1). So we can start finding maxR(L) from maxR(L-1). In this way out pointer R goes only forward.
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4. Best solution Notice that maxR(L)>=maxR(L-1). So we can start finding maxR(L) from maxR(L-1). In this way out pointer R goes only forward.

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Other examples There are 2 sorted arrays of integers: A and B. Count pairs (i, j) such that A[i]=B[j]. There are N points on circle. Find two points...
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Other examples There are 2 sorted arrays of integers: A and B. Count pairs (i, j) such that A[i]=B[j]. There are N points on circle. Find two points such that distance between them is maximal. There are N points on circle. Find three points such that area of triangle is maximal.

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Additional links and home task Article about two pointers method Discussion on codeforces Contest
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Additional links and home task Article about two pointers method Discussion on codeforces Contest

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