Can, the recurrence for this question be :
dp[i][j] = for all x min(dp[i][x] + dp[x][j] + 2 ) if arr[i] not equal to arr[j] else
min(dp[i][x] + dp[x][j] + 1 ).The intution behind this approach is trying all possible conversions i.e. the ith element to all colors from (1 to i-1).
Flood Fill , Different states for storing in DP
Hi @Vishal_234 ur solution seems right , but this solution is not feasible at all as in every state (i,j) u are traversing i to j making your complexity O(n^3) . So ,its better to stick our original approach i.e
for the case when arr[i]=arr[j] then dp[i][j] = 1+dp[i-1][j-1]
else dp[i][j]=1+min(dp[i+1][j],dp[i][j-1])
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