sethuraj.32
BAN USER
We can use divide and conquer approach.
Let h[] be the array and kth element be the minimum element in h[] from index i to j
max_area = max(max_area(i, k-1), max_area(k+1, j), h[k]*(j-i))
i will be set to 0 and j will be set to array length initially
Hash the array.
Iterate the array and check if x-a[i] exists in the hash.
Corner case is if x=4 and a[i]=2 and no other two is there in the array.
I think this code will work
public class FindCombi {
public static void main(String[] args) {
int s = 5;
int n = 3;
helper(n, s);
}
private static void helper(int n, int s) {
int[] a = new int[n];
finAllCombi(a, n, s, 0, 0);
}
private static void finAllCombi(int[] a, int n, int s, int sum, int i) {
if(i+1==n) {
a[i] = s - sum;
print(a);
return;
}
for(int m=0; m<=s-sum; m++) {
a[i] = m;
finAllCombi(a, n, s, sum+m, i+1);
}
}
private static void print(int[] a) {
String st = "";
for(int i=0; i<a.length; i++) {
st = st + a[i] ;
if(i+1 != a.length) {
st = st + "+";
}
}
System.out.println(st);
}
}
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- sethuraj.32 November 04, 2012package algo.solutions; public class Sort0s1s { public void sort(int[] a) { if(a == null || a.length == 1) { return; } int i=0; int j = a.length -1; while(true) { while(i < a.length && a[i] == 0) { i++; } while(j >= 0 && a[j] == 1) { j--; } if(i>j) { return; } swap(a,i,j); } } private static void swap(int[] a, int i, int j) { int tmp = a[i]; a[i] = a[j]; a[j] = tmp; } }