You have N stones in a row, and would like to create from them a pyramid. This pyramid should be constructed such that the height of each stone increases by one until reaching the tallest stone, after which the heights decrease by one. In addition, the start and end stones of the pyramid should each be one stone high.
You can change the height of any stone by paying a cost of 1 unit to lower its height by 1, as many times as necessary. Given this information, determine the lowest cost method to produce this pyramid.
For example, given the stones [1, 1, 3, 3, 2, 1], the optimal solution is to pay 2 to create [0, 1, 2, 3, 2, 1].
Solution.
If we are given a max height and its index in the stones array, then it is easy to get the cost in O(N) time.
To get this max height and its index, we do the following.
1. build an increasing array of interval 1 from left to right, starting at 1.
2. build an increasing array of interval 1 from right to left, starting at 1.
3. get the min of the above two arrays at the same indices and find the max of all these mins. This is the max possible height.
Both the runtime and space complexity are O(N).
public class ConstructPyramid {
public static int minCost(int[] stones) {
int n = stones.length;
int[] left = new int[n], right = new int[n];
left[0] = 1;
for(int i = 1; i < n; i++) {
left[i] = Math.min(left[i - 1] + 1, stones[i]);
}
right[n - 1] = 1;
for(int i = n - 2; i >= 0; i--) {
right[i] = Math.min(right[i + 1] + 1, stones[i]);
}
int maxHeight = 0, centerIdx = -1;
for(int i = 0; i < n; i++) {
int currHeight = Math.min(left[i], right[i]);
if(currHeight > maxHeight) {
maxHeight = currHeight;
centerIdx = i;
}
}
return getCost(stones, maxHeight, centerIdx);
}
private static int getCost(int[] stones, int maxHeight, int centerIdx) {
int cost = 0, h = maxHeight;
for(int i = centerIdx; i >=0; i--) {
if(h > stones[i]) {
return -1;
}
cost += stones[i] - h;
if(h > 0) {
h--;
}
}
h = maxHeight - 1;
for(int i = centerIdx + 1; i < stones.length; i++) {
if(h > stones[i]) {
return -1;
}
cost += stones[i] - h;
if(h > 0) {
h--;
}
}
return cost;
}
public static void main(String[] args) {
int[] stones = {1, 1, 3, 3, 2, 1};
System.out.println(minCost(stones));
}
}