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| 1 | +We will cut the array in 4 subarrays with 3 endpoints - i, j, k |
| 2 | + |
| 3 | +Let us fix j and find the best i and best k for every j. |
| 4 | + |
| 5 | +We will then choose the j that minimizes the difference of the 4 segments. |
| 6 | + |
| 7 | +----- |
| 8 | + |
| 9 | +Now, let us discuss how to find the best i for a given j ? |
| 10 | +We want to know the best way to split [1, j] into [1, i] and [i + 1, j] such that the difference in their sums is minimized. |
| 11 | + |
| 12 | +We can do this by binary searching for i for every j. |
| 13 | + |
| 14 | +We can also make another beautiful observation. |
| 15 | +Let Answer(j) denote the best value of i for j |
| 16 | + |
| 17 | +Then, Answer(j + 1) >= Answer(j) |
| 18 | + |
| 19 | +We can now use two pointers and use this to find the best i for each j by starting from the best i for the previous j ! |
| 20 | + |
| 21 | +This is O(n) as every element is touched at most twice. |
| 22 | + |
| 23 | + |
| 24 | +----- |
| 25 | + |
| 26 | +We will do a similar thing for the suffix in splitting [j + 1, n] into [j + 1, k - 1] and [k, n] |
| 27 | + |
| 28 | +----- |
| 29 | + |
| 30 | +#include <iostream> |
| 31 | +#include <vector> |
| 32 | +#include <algorithm> |
| 33 | + |
| 34 | +using namespace std; |
| 35 | + |
| 36 | +long long find_difference(long long a, long long b, long long c, long long d) |
| 37 | +{ |
| 38 | + long long minimum = min(min(a, b), min(c, d)); |
| 39 | + long long maximum = max(max(a, b), max(c, d)); |
| 40 | + |
| 41 | + return maximum - minimum; |
| 42 | +} |
| 43 | + |
| 44 | +int main() |
| 45 | +{ |
| 46 | + int no_of_elements; |
| 47 | + cin >> no_of_elements; |
| 48 | + |
| 49 | + vector <int> A(no_of_elements + 1); |
| 50 | + for(int i = 1; i <= no_of_elements; i++) |
| 51 | + { |
| 52 | + cin >> A[i]; |
| 53 | + } |
| 54 | + |
| 55 | + vector <long long> sum(no_of_elements + 1); |
| 56 | + for(int i = 1; i <= no_of_elements; i++) |
| 57 | + { |
| 58 | + sum[i] = sum[i - 1] + A[i]; |
| 59 | + } |
| 60 | + |
| 61 | + vector <long long> best_P(no_of_elements + 1), best_Q(no_of_elements + 1); |
| 62 | + for(int i = 1, j = 2; j <= no_of_elements; j++) |
| 63 | + { |
| 64 | + //[1, i], [i + 1, j] |
| 65 | + int break_point = i; |
| 66 | + for(break_point = i; break_point < j; break_point++) |
| 67 | + { |
| 68 | + long long segment_1 = sum[break_point], segment_2 = sum[j] - sum[break_point]; |
| 69 | + long long segment_1_before = sum[break_point - 1], segment_2_before = sum[j] - sum[break_point - 1]; |
| 70 | + //cout << "j = " << j << " Before " << segment_1_before << " " << segment_2_before << " Now " << segment_1 << " " << segment_2 << "\n"; |
| 71 | + |
| 72 | + if(abs(segment_2 - segment_1) <= abs(segment_1_before - segment_2_before)) |
| 73 | + { |
| 74 | + i = break_point; |
| 75 | + } |
| 76 | + else |
| 77 | + { |
| 78 | + break; |
| 79 | + } |
| 80 | + } |
| 81 | + |
| 82 | + best_P[j] = sum[i], best_Q[j] = sum[j] - sum[i]; |
| 83 | + } |
| 84 | + |
| 85 | + vector <long long> best_R(no_of_elements + 1), best_S(no_of_elements); |
| 86 | + for(int k = no_of_elements, j = no_of_elements - 1; j >= 1; j--) |
| 87 | + { |
| 88 | + //[j + 1, k - 1], [k, n] |
| 89 | + int break_point = k; |
| 90 | + for(break_point = k; break_point > j; break_point--) |
| 91 | + { |
| 92 | + long long segment_3 = sum[break_point - 1] - sum[j], segment_4 = sum[no_of_elements] - sum[break_point - 1]; |
| 93 | + long long segment_3_before = sum[break_point] - sum[j], segment_4_before = sum[no_of_elements] - sum[break_point]; |
| 94 | + |
| 95 | + if(abs(segment_3 - segment_4) <= abs(segment_3_before - segment_4_before)) |
| 96 | + { |
| 97 | + k = break_point; |
| 98 | + } |
| 99 | + else |
| 100 | + { |
| 101 | + break; |
| 102 | + } |
| 103 | + } |
| 104 | + |
| 105 | + best_R[j] = sum[k - 1] - sum[j], best_S[j] = sum[no_of_elements] - sum[k - 1]; |
| 106 | + } |
| 107 | + |
| 108 | + const long long oo = 1e18; |
| 109 | + long long answer = oo; |
| 110 | + for(int j = 2; j < no_of_elements; j++) |
| 111 | + { |
| 112 | + //cout << "j = " << j << "(" << best_P[j] << "," << best_Q[j] << "," << best_R[j] << "," << best_S[j] << ")\n"; |
| 113 | + answer = min(answer, find_difference(best_P[j], best_Q[j], best_R[j], best_S[j])); |
| 114 | + } |
| 115 | + |
| 116 | + cout << answer << "\n"; |
| 117 | + return 0; |
| 118 | +} |
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