i think the problem can also be solved by DP..
arr is the original array..
final is the array required..
for(i=0;i<n;i++) final[i]=arr[i];
for(i=2;i<=k;i++)
for(j=0;j<n-i+2;j++)
final[i]=max( final[j] , final[j+1]);
initial n-i+2 entries in the array final contains the answer..
time complexity (O(nk))...
correct me if i'm wrong anywhere...
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