Below is my C solution for the Leetcode problem candy. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 static inline void previous_kids_started_crying(int* kid, int continous_decrease_start, int continous_decrease_end, int* candy) { /* * To satisfy the condition that each child with higher rating than * it's neighbors must get more candies. We give all children in range * [continous_decrease_start, continous_decrease_end] an extra candy * if the he has candies lower or equal to the kid after it. * * Note that we do not bother the kid before it, since it is known * that the kid before it has more rating than this kid */ if (continous_decrease_start == -1) { exit(1); } for (int i = continous_decrease_end; i >= continous_decrease_start; i--) { if (kid[i] <= kid[i+1]) { int before = kid[i]; kid[i]++; int after = kid[i]; *candy += after - before;; } else { /* chain is broken. We found a satisfied non protesting kid the kids before it must also be non-protesting */ return; } } } int candy(int* kid, int n) { /* * Few things to take care of: * * 1. At a time take a look at 2 children. i.e. iterate the children * in window of 2. * * 2. Each time a candy is given, check if previous child has more * rating, since he will start crying. He/she will protest that the * kid in front of him has less rating than him/her and still got * more/equal candies than him/her. It might even trigger a chain * where the kid previous to the previous kid might also see this * changed state, and if he had a rating more then the kid after him. * then he will also start asking more candies. And the kid behind * that and so on. * * The algorithm runs in linear time, however, if the children are * arranged in decreasing order of rating, then at each iteration all * previous children will start crying. Hence, everybofy needs to be * given more candies. In that case it becomes quadratic * * Space required is constant. * * Best case: * Time: O(n) * Space: O(1) * * Worst Case: * Time: O(n^2) * Space: O(1) */ int candies = 0; int continous_decrease_from = -1; /* First child gets a candy, but save his rating first */ int previous_child_rating = kid[0]; kid[0] = 1; candies++; for (int i = 1; i < n; i++) { if (kid[i] > previous_child_rating) { /* Since this child has more rating than the previous child, this child gets 1 more candy than previous child */ previous_child_rating = kid[i]; kid[i] = kid[i-1]+1; candies += kid[i]; continous_decrease_from = -1; } else if (kid[i] == previous_child_rating){ /* Since this child has equal rating than the previous child, give him 1 candy */ previous_child_rating = kid[i]; kid[i] = 1; candies++; continous_decrease_from = -1; } else { /* Previous child has more rating. Give this child one candy, and after giving : if the previous child had less candy, then we must initiate a chain reaction to do justice to all previous kids to previous children the problem here is that if this child has less rating then his predecessor then he will also start crying. And, the predecesor of this child also has more score then he will also start crying. Hence, we must give candies to them as well to maintain rule. Hence, check if the rule is disturbed, */ if (continous_decrease_from == -1) { /* A period of continous decrease started */ /* Hence, give this child 1 candy and if the previous child will also get 1 candy if he gets */ continous_decrease_from = i-1; previous_child_rating = kid[i]; kid[i] = 1; if (kid[i-1] > kid[i]) { candies ++; } else { kid[i-1] = kid[i] + 1; candies += 2; } } else { /* Give this child a candy, but mind the children behind this child */ previous_child_rating = kid[i]; kid[i] = 1; candies++; previous_kids_started_crying(kid, continous_decrease_from, i-1, &candies); } } } return candies; }

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