How do I do this question?
Click here for problem statement
Not able to understand the editorial.
Thanks in advance.
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For any position i in the array, for what length “window” can that number act as strength ?
for example, the array is
3 1 4 2 3 0
Clearly, the 2 at 3rd position (0th based indexing) will act as strength for the windows of length 1,2 and 3 only. This is because,if you expand the window on either side, we have the number 1 ( <2 ) and number 0 ( <2) .
So, for each index i we need to find the largest index j < i such that a[j] < a[i]. Let’s store it in l[i].
For each index i, we need to find the largest index j > i such that a[j] < a[i]. Let’s store it in r[i].
Clearly, the maximum window which you can get is from l[i] + 1 to r[i] - 1, which has a length of r[i] - l[i] - 1. Therefore, each element can act as strength for window size from 1 to r[i] - l[i] - 1.
You can calculate arrays l and r using a stack and then solve the problem.
http://codeforces.com/contest/547/submission/19434836
There is one very simple method. You can easily find the answer for x = 1. Initialize the array, a[200005][200005]. Therefore, a[1][1] = a1, a[1][2] = a2, …, a[1][i] = ai, …, a[1][n] = an. In the loop you can store maximum strength in tempmax while assigning values to the array. print - tempmax
tempmax = 0
Now for x = 2. a[2][2] = min(a[1][1], a[1][2]), a[2][3] = min(a[1][2], a[1][3]), …, a[2][i] = min(a[1][i-1], a[1][i]), …, a[2][n] = min(a[1][n-1], a[1][n]). print - tempmax
…
tempmax = 0
For x = i. a[x][x] = min(a[x-1][x-1], a[x-1][x]), a[x][x+1] = min(a[x-1][x], a[x-1][x+1]), …, a[x][x+i] = min(a[x-1][x+i-1], a[x-1][x+i]), …, a[x][n] = min(a[x-1][n-1], a[x-1][n]). print - tempmax
…
tempmax = 0
For x = n a[n][n] = min(a[n-1][n-1], a[n-1][n]) if(a[n][n] > tempmax) tempmax = a[n][n]. printf - tempmax
Now comes the question that logic is perfectly fine, but we can’t declare such a huge array. Again, there is a simple solution for that. The method to implement it using 1D array is as follows:
declare array a[200005]
tempmax = 0;
x = 1:
for(i=n to 1){
a[i] = ai;
if(a[i]>tempmax) tempmax = a[i];
}
printf("%d ", tempmax);
tempmax = 0;
x = 2:
for(i=n to 2){
a[i] = min(a[i], a[i-1]);
if(a[i]>tempmax) tempmax = a[i];
}
printf("%d ", tempmax);
…
tempmax = 0;
x = j:
for(i=n to j){
a[i] = min(a[i], a[i-1]);
if(a[i]>tempmax) tempmax = a[i];
}
printf("%d ", tempmax);
…
tempmax = 0;
x = n:
for(i=n to n){
a[i] = min(a[i], a[i-1]);
if(a[i]>tempmax) tempmax = a[i];
}
printf("%d ", tempmax);
Now I guess you got the notion about how to approach the question. The final code which you can implement is as follows:
//min function
int n, a[200005], tempmax, i, j;
//read the inputs
tempmax = 0;
for(i=1;i<=n;i++){
a[i] = ai;
if(a[i] > tempmax)
tempmax = a[i];
}
printf("%d ", tempmax);
for(j=2;j<=n;j++){
tempmax=0;
for(i=n;i>=j;i--){
a[i] = min(a[i-1], a[i]);
if(a[i] > tempmax)
tempmax = a[i];
}
printf("tempmax ");
}
But this method isn’t sufficient as it will give TLE. Please implement it using BIT.