hey!! I watched your videos. Nice explanation. Good job done :) I have a doubt in problem E. You figured out that cyclic dp will create a problem as the recursion will catch itself in loop. So, we should use dijikstra, to find minimum distance from source(0) to destination(n), by considering dp states. So, the solution Time complexity will be : m*(g+r)*log(m*(g+r)), which will cause TLE. So, how to optimise it?

Want to add: topological sort can only be used with directed acyclic graph(DAG).

Let a[1], a[2].. a[n] be the elements in the array. Let 1 <= i < j < k < l....< n be the k-1 indices we choose for dividing the array into k segments. So, cost of division = (a[i] — a[1]) + (a[j] — a[i+1]) + .... + (a[k] — a[j+1]) + (a[n] — a[k+1]), on rearranging : (a[n] — a[1]) — ((a[i+1] — a[i]) + (a[j+1] — a[j]) +.....+ (a[k+1] — a[k])). So, we will store the difference of a[i+1]-a[i] for each 1<=i<n , sort them in decreasing order and take first k-1 differences and finally subtract from (a[n]-a[1]).

f(b, c) == even no of changes

let number of 1 in string b = x1, number of 1 in string c = x2, number of common 1 in both string = x,

so, number of changes = x1 + x2 — 2*x, which should be even. /** 2*x is always even **/, x1 + x2 should be even , it implies x1 and x2 should be both even or both odd, hence should have same parity