Frog 2 at codeer dp

#1

for(ll j=1; j<=k; j++) {
if(i-j>=0) {
dp[i] = min(dp[i], dp[i-j] + llabs(h[i-j]-h[i]));
}

In this dp provlem I camnnot uder stand significance of this line
dp[i-j] + llabs(h[i-j]-h[i]));

#2

hello @Somasree

here we r at i and we have take j jump.
that means its previous location will be i-j.

dp[i]=cost to reach i-j + cost of jump from i-j to i.
dp[i]=dp[i-]+abs(h[i-j]-h[i]));

#3

ok understood.thank u

#4

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